Fhe  D.  Van  No^and  Company 
Lend  this  book  to  be  sold  to  the  Pubhc 

rt  the  advertised  price,  and  supply     to 
t  T»de  on  terms  wHch  will  not  allow 

of  reduction. 


SOLUTION  OF  RAILROAD 

PROBLEMS  BY  THE 

SLIDE  RULE 


BY 

E.  R.  CARY,  C.E. 

Professor  of  Railroad  Engineering  and  Geodesy,  Rens- 

selaer  Polytechnic  Institute,  Troy,  N.Y. 

Member  American  Society 

Civil  Engineers 


NEW  YORK 

D.  VAN  NOSTRAND  COMPANY 

25  PARK  PLACE 

1913 


COPYRIGHT,  1913, 

BY 
D.  VAN  NOSTRAND  COMPANY 


Stanbope  iprcss 

F.    H.GILSON   COMPANY 
BOSTON,  U.S.A. 


PREFACE 


THE  ease  and  rapidity  of  solving  problems 
in  Railroad  Curves  by  the  use  of  the  slide  rule 
led  the  author  to  develop  a  set  of  problems  for 
class-room  work.  The  object  of  this  book  is 
to  present  similar  problems  for  the  convenience 
of  students  who  have  studied  Railroad  Curves 
and  the  Theory  of  the  Slide  Rule. 

Where  it  is  possible  the  solution  of  the 
problems  should  be  made  by  the  use  of  the 
C  and  D  scales  of  the  Mannheim  slide  rule  on 
account  of  their  greater  precision. 

The  notation  used  in  Allen's  "  Railroad 
Curves  and  Earthwork"  is  used  thruout  this 
discussion. 

The  discussion  of  the  slide  rule  is  from 
Professor  C.  W.  Crockett's  article  in  the 
Polytechnic  of  May  2,  1910. 

The  discussion  of  the  slide  rule,  the  develop- 
ment of  the  equations  used  and  the  discussion 
of  the  easement  curve  have  been  added  to 
make  this  book  of  more  general  interest. 

E.  R.  GARY. 

TROY,  N.  Y. 
March  1,  1913. 

iii 


288606 


TABLE  OF  CONTENTS 


CHAPTER  I 
THE  SLIDE  RULE 

AET.  PAGE 

1.  The  Construction  and  Setting  of  the  Slide 

Rule 1 

2.  The  Mannheim  Slide  Rule 4 

CHAPTER  II 
SIMPLE  CURVES 

3.  The  Center  Line 5 

4.  The  Degree  of  a  Curve 5 

5.  The  Decimal  Point 7 

6.  The  Functions  of  a  Curve 7 

7.  The  Middle  Ordinate 10 

8.  The  Tangent  Offset 12 

9.  A  Curve  laid  out  by  a  Tape 13 

10.  Offsets  from  a  Chord 15 

11.  Deflection  Angles 16 

12.  Offsets  from  a  Tangent 19 

13.  Special  Problems 21 

14.  Other  Special  Problems 28 

CHAPTER   III 
COMPOUND  CURVES 

16.  Compound  and  Reversed  Curves 31 

17.  Method  of  Solution  by  Coordinates 33 

v 


VI  TABLE   OF   CONTENTS 


CHAPTER  IV 
VERTICAL  CURVES 

ART.  PAGE 

18.  The  Vertical  Curve ...  48 


CHAPTER  V 

TURNOUTS 

19.  The  Split  Switch  Turnout 51 

20.  The  Turnout  from  a  Curve 54 

21.  The  Stub  Switch  Turnout 57 

22.  The   Approximate   Method   for   a  Turnout 

from  a  Curve 58 

23.  The  Connecting  Curve 62 

24.  A  Connection  between  Parallel  Tracks 63 

25.  A  Connection  between  Concentric  Curves ...  64 

26.  The  «  Y  "  Curve 70 

CHAPTER  VI 

THE  EASEMENT  CURVE 

27.  The  Easement  Curve 75 

28.  The  True  Spiral 78 

29.  The  Spiral  Angle 79 

30.  The  Deflection  Angles  for  a  Spiral 80 

31.  The  Offset  from  the  Tangent 81 

32.  The    Deflections    from   the    Tangent    thru 

the  P.S 82 

33.  The  Value  of  p 83 

34.  The  Value  of  i 83 

35.  The  Value  of  q 85 

36.  The  Value  of  Ta 87 

37.  To  find  the  Tangent  at  any  Point 87 


TABLE    OF   CONTENTS  Vll 


38.  The  Deflections  from  any  Point  on  the  Spiral  88 

39.  The  Methods  of  Laying  out,  Spirals 89 

40.  The  Offset  Method 91 

41.  The  Spiral  between  the  Branches  of  a  Com- 

pound Curve 93 

42.  The  Spiral  Tables 94 

44.   Spirals  not  given  in  the  Tables 95 

46.  The  Solution  of  a  Spiral  by  the  Slide  Rule. .  104 

47.  The    Equation    for    Deflections,    using    the 

Slide  Rule 112 

48.  The  Diagram  for  e 113 

49.  The  Diagram  for  lc 115 

CHAPTER  VII 

EARTHWORK 

50.  Cross-sections 117 

51.  Form  of  Notes  for  Cross-sections 118 

52.  Change  from  Cut  to  Fill 119 

53.  A  L^vel  Section 120 

54.  A  Three-level  Section 121 

55.  The  Methods  of  Computation 123 

56.  The  Prismoidal  Correction 123 

57.  The  Formulas  for  Slide  Rule  Computations.  126 

PROBLEMS 

PBOB. 

1.  To  find  the  Radius 6 

2.  To  find  the  Tangent  Distance 8 

3.  To  find  the  External  Distance 9 

4.  To  find  the  Middle  Ordinate 10 

5.  To  find  the  Long  Chord 10 


Vlll  TABLE    OF   CONTENTS 


6.  To  find  the  Middle  Ordinate 12 

7.  To  find  the  Tangent  Offset 12 

8.  To  find  the  Offset  from  any  Point  on  a  Chord  16 

9.  To  find  the  Deflections 17 

10.  To  locate  a  Curve  by  Offsets  from  the  Tan- 

gent   19 

11.  To  change  the  Ending  Tangent 24 

12.  To  change  the  Ending  Tangent 25 

13.  To  change  the  Ending  Tangent 26 

14.  Special  Problem  for  a  Simple  Curve „  28 

15.  Tangent  Distances  for  a  Compound  Curve. .  32 

16.  Compound  Curve 33 

17.  Compound  Curve 36 

18.  Compound  Curve 39 

19.  Changing  a  Simple  to  a  Compound  Curve. . .  41 

20.  Three-centered  Compound  Curve 43 

21.  A  Reversed  Curve 45 

22.  A  Reversed  Curve . . .  46 

23.  A  Vertical  Curve 48 

24.  A  Split  Switch  Turnout 53 

25.  A  Turnout  from  a  Curve 60 

26.  A  Connecting  Curve 62 

27.  A  Connection  between  Parallel  Tracks 64 

28.  A  Turnout  and  Connecting  Curve  for  Con- 

centric Curved  Tracks 66 

29.  Similar  to  Prob.  28 68 

30.  A  "Y"  Curve 70 

31.  A  "Y"  Curve 72 

32.  Deflections  for  a  Spiral,  using  the  Tables 97 

33.  Deflections  for  a  Spiral,  using  the  Slide  Rule.  104 

34.  Earthwork,  using  the  Slide  Rule 128 

35.  Earthwork,  using  the  Slide  Rule 128 


TABLE   OF   CONTENTS  IX 

PROS.  PAGE 

36.  Earthwork,  using  the  Slide  Rule. .  „ 129 

37.  Earthwork,  using  the  Slide  Rule 130 

38.  Earthwork,  using  the  Slide  Rule 130 

39.  Earthwork,  using  the  Slide  Rule 131 

DIAGRAMS 

For  e 114 

For  lc 116 

TABLES 

For  Deflection  Angles  for  Spirals ..........     100-103 

Constants 132 

Formulas  for  Triangles 132 

Trigonometric  Formulas "  133 

Trigonometric  Series 134 

Simple  Curve  Formulas 134 

Vertical  Curve  Formulas 135 

Turnout  Formulas 135 

Spiral  Formulas 136 

Earthwork  Formulas 136 


THE  SOLUTION 

OF 

RAILROAD  PROBLEMS 

BY 

THE  SLIDE  RULE 


THIS  discussion  will  be  given  under  the  following 
heads:  The  Slide  Rule,  Problems  in  Simple  Curves, 
Problems  in  Compound  Curves,  The  Vertical  Curve, 
Turnouts,  The  Easement  Curve,  and  Problems  in 
Earthwork.  

CHAPTER  I 

THE  SLIDE  RULE 

i.  The  Mannheim  and  Carpenter  rules  are 
the  most  common  types  of  the  slide  rule.  The 
face  of  either  bears  two  kinds  of  scales,  on  one 
of  which,  for  the  ordinary  ten-inch  rule,  the 
distance  from  1  to  1  is  about  5  inches,  while 
on  the  other  the  distance  from  1  to  1  is  about 
10  inches,  the  latter  distance  being  exactly 
double  the  former. 

Let  us  call  the  shorter  scale  the  single  scale 
and  the  longer  scale  the  double  scale;  on  the 
1 


%  THE-    SLIDE   RULE 

Mannheim  rule,  the  single  scale  is  called  A  or 
B  and  the  double  scale  is  C  or  Z>;  on  the  Car- 
penter rule,  the  single  scale  is  called  A  or  B, 
and  the  double  scale  is  called  C.  Note  that 
the  double  scale  is  the  long  scale. 

Let  a  and  6  be  constant  numbers  and  x  a 
third  number  to  which  different  values  are 
to  be  assigned,  and  suppose  we  wish  to  find 
the  values  of  y  in  any  one  of  the  following 
expressions: 

ax          ax2          A  fax  t  /ax2 

y  =  j>y  =  -v>y=-\T'y==\-v' 

ab  o62  .  fab  .  /ah* 

y=lt>y=-^'y=-\^'y  =  \l?' 

or  of  any  similar  expression  containing  two 
factors  in  the  numerator  and  one  in  the  de- 
nominator, any  one  or  more  being  squared, 
a  being  a  constant  factor  in  the  numerator. 

The  method  of  setting  the  instrument  is 
as  follows: 

1°.  If  £  is  in  the  numerator,  use  the  slide 
direct. 

If  x  is  in  the  denominator,  use  the  slide 
inverted. 

2°.  The  numbers  a,  &,  x  and  y  are  found 
on  the  instrument  in  the  order  named;  a  on 


THE    SLIDE   RULE  O 

the  rule,  6  on  the  slide,  x  on  the  slide,  y  on 
the  rule;  that  is,  we  look  on  the  rule  and  slide 
in  the  following  order:  Rule,  slide,  slide,  rule. 
Or  we  could  find  a  on  the  slide,  b  on  the  rule, 
x  on  the  rule,  y  on  the  slide;  that  is,  we  look 
on  the  rule  and  slide  in  this  order,  Slide,  rule, 
rule,  slide. 

3°.  If  a,  b  or  x  is  raised  to  the  first  power, 
we  must  find  it  on  a  single  scale;  if  squared, 
on  a  double  scale. 

4°.  If  y  is  a  first  power,  we  must  find  it 
on  a  single  scale;  if  a  square  root,  on  a  double 
scale. 

ax2 
Thus  to  find  the  value  of  y  =  -p-  with  the 

Mannheim  rule,  we  may  proceed  in  two  ways, 
using  the  slide  direct,  since  x  is  in  the  numerator: 

(a)  Find  a  on  the  single  scale  on  the  rule, 
and  opposite  it  set  b  on  the  double  scale  on 
the  slide;    find   x  on  the  double  scale  on  the 
slide,  and  opposite  it  read  the  result  y  on  the 
single  scale  on  the  rule.     Or: 

(b)  Find  a  on  the  single  scale  on  the  slide, 
and  opposite  it  set  b,  found  on  the  double  scale 
on  the  rule;   find  x  on  the  double  scale  on  the 
rule,  and  opposite  it  read  the  result  y  on  the 
single  scale  on  the  slide. 


4  THE   SLIDE   RULE 

The  same  problem  may  be  solved  with  the 
Carpenter  or  Thacher  rule,  but  only  by  the 
second  method,  since  the  slide  does  not  bear  a 
double  scale;  this  limitation  is  not  objection- 
able, however,  except  in  continued  operations, 
where  it  is  desirable  to  use  the  runner. 

/ax2 
Had  we  wished  to  find  the  value  of  y  =  y  -j^> 

the  settings  on  the  Mannheim  rule  would  have 
been  the  same  as  before,  but  the  result  y  would 
have  been  on  the  double  scale  instead  of  on  the 
single  scale.  The  Carpenter  or  Thacher  rule 
cannot  solve  this  problem  without  first  reading 

ax2 

on  the  slide  the  value  of  -r^-,  and  then  deter- 
mining its  square  root. 

2.  The  Mannheim  slide  rule  is  used  in  the 
solution  of  the  problems  given  herein.  The 
scales  of  this  rule  are  as  follows:  The  upper 
scale  of  the  rule,  A;  the  upper  scale  of  the 
slide,  E\  the  lower  scale  of  the  slide,  C;  the 
lower  scale  of  the  rule,  D.  On  the  back  of 
the  slide  are  the  sine  and  tangent  scales,  and 
when  these  are  to  be  used  the  slide  must  be 
reversed,  i.e.,  the  slide  changed  in  the  rule  so 
that  the  back  face  is  brought  to  the  front. 


CHAPTER  II 


SIMPLE  CURVES 

3.  The  center  line  of  a  railroad  track  con- 
sists of  tangents  and  curves   (circular  arcs), 
and,  in  modern  practice,  also  of  some  form  of 
easement  curve  connecting  them. 

4.  In  this  country  a  curve  is  designated  by 
its  degree.     The  degree  of  a  curve  is  the  cen- 
tral  angle   subtended 

by  a  chord  of  one, 
hundred  feet.  If  the 
metric  system  is  used, 
the  degree  of  a  curve 
may  be  defined  as  the 
central  angle  sub- 
tended by  a  chord  of 
ten  meters.  In  almost 
all  other  countries  a 
curve  is  designated  by  its  radius. 
From  Fig.  1, 

R  sin  \  D  =  50; 
50 


R  = 


sin  J  D 
5 


6  THE    SLIDE   RULE 

As  D  is  usually  a  small  angle, 
50 


R  = 


iZ>sml' 

looJL, 


-4-,  =  3438; 
sinl' 

100  X  3438 
«  =       -p-  -•  •     •    •     (1) 

In  Equation  (1),  D  is  expressed  in  minutes 
and  R  in  feet.     For  metric  curves  R  in  meters 
_  10  X  3438 
D' 

Problem  1.  Find  the  radius  of  a  6°  30'  curve. 
To  solve  by  the  slide  rule,  opposite  3438  on 
the  D  scale  set  D  in  minutes  on  the  C  scale, 
opposite  the  index  of  the  C  scale  read  the 
result  on  the  D  scale  for  R  in  feet. 

By  inverting  the  slide  and  setting  3438  on 
the  C  scale  opposite  the  index  of  the  D  scale, 
the  radius  of  any  degree  of  curvature  may  be 
found  on  the  D  scale  opposite  the  degree  of 
the  curve  in  minutes  on  the  inverted  C  scale. 

For  the  above  problem  the  setting  is  as 
follows: 

Opposite  3438  on  the  D  scale  set  390  on  the 


SIMPLE    CURVES  7 

C  scale  and  opposite  the  right  index  of  the  C 
scale  read  882  on  the  D  scale,  or  R  =  882  feet. 
Except  for  odd  minute  curves  the  approximate 

,         ,      D      5730      .  ,   ,.       , 

formula,  R  =  — ~- ,  gives  an  easy  solution  by 

the  slide  rule.  In  this  formula  D  is  in  degrees. 
If  this  formula  is  figured  exactly  and  T£o  of  D 
is  added  to  the  value  found  for  R,  the  result 
is  practically  a  precise  value  of  R. 

5.  To  find  the  decimal  point,  always  figure 
the  result  roughly  by  mental  arithmetic,  e.g., 
in  the  above  problem  use  3900  in  place  of 
3438  and  the  result  is  1000,  showing  that  the 
correct  result  is  about   1000.     Hence,   when 
the  figures  882  are  obtained  by  the  slide  rule, 
the  decimal  point  is  placed  after  the  third 
figure  for  the  value  of  R. 

6.  The  functions  of  a  curve  are:    T7,  the 
tangent  distance;    E,  the  external  distance; 
M,  the  middle  ordinate;    (7,  the  long  chord; 
P.O.,  the  beginning  of  the  curve;    P.T.,  the 
ending  of  the  curve;    and  /  the  central  angle 
of  the  curve. 

In  Fig.  2,          T  =  AV  =  VB. 
E  =  VF. 
M  =  FG. 
C  =  AB. 
I  =  ACB  =  BVX. 


8 


THE   SLIDE   RULE 


From   Fig.    2   the   following   formulas  are 
readily  derived: 

r  =  #tani7  ........  (2) 

E  =  R  (sec  I  I  -  1)  =  R  exsec  i  7.  (3) 
exsec  J  7  =  tan  J  7  tan  }  7, 

or    #  =  R  tan  £  7  tan  i  7  .  .  .  .  (4) 

Af  =  JB-72.cosf  J  =  Bversi/     .  (5) 

vers  i  7  =  2  sin2  i  7  =  sin  ^  7  tan  J  7, 

or       Af  =  jRsini7tani7  .....  (6) 

C  =  2  E  sin  i  7  .......  (7) 


Fig.  2 

Problem  2.   Find  the  tangent  distance  for  a 
6°  12'  curve  with  a  central  angle  of  30°  24'. 

T  =  R  tan  i  7  =  E  tan  15°  12'. 

fi,  found  as  in  Problem  1,  =  925  feet. 

T  =  925  tan  15°  12'  =  251.5  feet. 


SIMPLE   CURVES  9 

The  setting  for  T  is  as  follows:  Using  the 
slide  reversed,  opposite  925  on  the  D  scale  set 
the  right  index  of  the  T  scale  and  opposite 
15°  12'  on  the  T  scale  read  251.5  on  the  D  scale. 
In  figuring  roughly  for  the  decimal  point  use 
the  sin  or  tan  of  1°  =  0.0175  and  assume  that 
the  tangents  and  sines  vary  as  the  angles,  i.e., 
the  sin  of  15°  is  15  times  the  sin  of  1°.  This 
is  only  approximately  true  for  angles  less 
than  30°. 

Problem  3.  Find  the  external  distance  for 
a  6°  12'  curve  with  a  central  angle  of  30°  24'. 

E  =  R  tan  £  7,  tan  J  / 

=  R  tan  15°  12',  tan  7°  36'. 
R,  the  same  as  in  Problem  2,  is  925  feet. 
E  =  925  tan  15°  12',  tan  7°  36'  =  33.5  feet. 

The  setting  for  E  is  as  follows:  Using  the 
slide  reversed,  set  the  right  index  of  the  T  scale 
opposite  925  on  the  D  scale,  bring  the  runner 
to  15°  12'  on  the  T  scale,  then  bring  the  left 
index  of  the  T  scale  under  the  runner,  and 
opposite  7°  36'  on  the  T  scale  read  the  result 
33.5  on  the  D  scale.  The  position  of  the 
decimal  point  is  found  as  indicated  in  Prob- 
lem 2. 


10  THE    SLIDE   RULE 

Problem  4.   Find  the  middle  ordinate  for  a 
6°  12'  curve  with  a  central  angle  of  30°  24'. 
M  =  R  sin  ^  7  tan  J  /. 
j?,  the  same  as  in  Problem  2,  is  925  feet. 
M  =  925  sin  15°  12'  tan  7°  36'  =  32.4  feet. 

The  setting  for  M  is  as  follows:  Using  the 
slide  reversed,  opposite  925  on  the  D  scale,  set 
the  right  index  of  the  T  scale  and  opposite 
7°  36'  on  the  T  scale  read  1235  on  the  D  scale. 
Opposite  1235  on  the  A  scal6  set  the  right  index 
of  the  S  scale  and  opposite  15°  12'  on  the  S 
scale  read  32.4  on  the  A  scale.  The  position 
of  the  decimal  point  is  found  as  indicated  in 
Problem  2. 

Problem  5.  Find  the  length  of  the  long 
chord  of  a  6°  12'  curve  with  a  central  angle  of 
30°  24'. 

jffi,  found  as  above,  =  925  feet. 

C  =  2  R  sin^  =  2  R  sin  15°  12'  =  2  X  242  = 

484  feet. 

Setting  for  925  sin  15°  12'  is  similar  to  that 
given  in  Problem  4. 

7.  In  bending  rails  it  is  convenient  to  use 
the  middle  ordinate  for  a  given  length  of 
chord  to  determine  the  degree  of  the  curve  to 
which  the  rail  is  bent.  The  middle  ordinate 


SIMPLE   CURVES 


11 


may  be  expressed  in  terms  of  the  length  of  the 
chord  and  the  radius  or  in  terms  of  the  length 
of  the  chord  and  the  degree  of  the  curve. 
These  formulas  are  derived  as  follows: 


In  Fig.  3,  AB  is  the  chord,  c,  and  EF  is  its 
middle  ordinate.     G  is  at  the  middle  of  AE. 

AE  =  AFj  approximately. 


—  O~E>  •     •     •     •     (8) 


Then     EF=AG 


\c=R 
M      ic 


or 


R  = 


or 

5730 
D    ' 


8  X  5730      45,840 


12  THE    SLIDE   RULE 

Problem  6.   Find  the  middle  ordinate  of  a 
chord  84  feet  long  on  a  6°  12'  curve. 

_     c*D     _  6.2 

~  45,840  ~          *  45,840  = 

The  setting  for  M  is  as  follows:  Opposite  84 
on  the  D  scale  set  4584  on  the  right  B  scale, 
E    opposite  62  on  the  left  B 
scale   read  953   on  the   A 
scale.     Figure  roughly  for 
the  decimal  point. 

8.   In  Fig.  4,  EB  is  the 
tangent   offset,   a,   and   its 
value  in  terms  of  chord,  c, 
c  and  the  radius  is  found  as 

Fis-  4  follows: 

Triangles  AEB  and  ACF  are  similar,  hence 

EB  _  AF 
AB~AC* 
but  AC  =  R,  AF  =  J  c. 

Then        -  =  ^     or     a  =  ~ .     .    ,  f  .     (9) 


Problem  7.   Find  the  tangent  offset  for  a 
6°  12'  curve  for  a  chord  of  110  feet. 

c2 


SIMPLE    CURVES 


13 


a  = 


R,  found  as  above,  =  925  feet. 
HO2         0.5  X  HO2 


=  6.55  feet. 


2  X  925  925 

Setting  for  A  is:  Opposite  5  on  the  A  scale 
set  925  on  the  B  scale  and  opposite  11  on  the 
C  scale  read  655  on  the  A  scale. 

9.  The  tangent  offset  may  be  used  to  lay 
out  a  curve  by  the  use  of  a  tape  only.  In 
Fig.  5,  if  A  is  at  station  16  +  40,  then 

602D 


A  17  =  60  feet,  and  E  17  =         = 


11,460* 


All  - 


Ell 


2A17 
From  this  AE  may  be  found. 


Fig.  5 

It  is  assumed  that  the  direction  of  the  line 
TA  is  established  on  the  ground  and  that  A 


14  THE    SLIDE   RULE 

is  located.  Then  AE  may  be  measured  out 
and  E  located.  Then  measuring  off  A 17 
from  A  and  E  17  from  E}  their  intersection 
may  be  found  and  17  located  as  a  point  on  the 
curve.  Point  16'  may  be  located  in  a  manner 
similar  to  locating  17.  Both  16'  and  17  are 
on  the  curve.  By  prolonging  16',  17, 100  feet 
from  17  to  (?,  measuring  G 18  =  2  aioo  (where 
aioo  is  the  tangent  offset  for  a  chord  of  100 
feet),  and  measuring  100  feet  from  17,  the 
intersection  of  these  measurements  locates 
station  18.  In  a  similar  manner  station  19 
may  be  located  from  the  prolongation  of  17, 18. 


18  H  is  aioo.    18,19  -  19  #  = 


200 


From  this  19  H  may  be  found.  Then  by  meas- 
uring 18  H  from  18  and  19  ff  from  19,  H  may 
be  located  and  the  line  19  H  is  tangent  to  the 
curve  at  station  19.  Then  station  19  +  30 
may  be  located  from  this  tangent  in  a  manner 
similar  to  locating  station  17  from  the  tangent 
thru  A. 


302 


19  K  =  77^5 .     30  minus  the  distance  from 
Zn, 


19  +  30  to  K  equals  .     From  these  dis- 


SIMPLE   CURVES 


15 


tances  K  may  be  located,  and  K  joined  with 
station  19  +  30  gives  the  direction  of  the 
tangent  ahead  of  the  curve. 

10.  A  curve  may  be  located  by  offsets  from 
a  long  chord.  The  method  of  finding  the  off- 
sets is  as  follows: 


F    K 


Fig.  6 

In  Fig.  6,      JK  =  LK  -  LJ. 
LK  =  EF. 

Let  AB  =  c,    and  W£  =  6. 

Then  </#,  the  offset  at  K,  =  /=  -  ^^^ 


or 


2R 


2R 


.    .     .     .     (10) 


16  THE    SLIDE    RULE 

By  Equation  (10)  the  offset  to  the  curve  from 
any  point  on  the  chord  is  equal  to  the  product 
of  the  segments  of  the  chord  divided  by  twice 
the  radius  of  the  curve. 

Problem  8.  Find  the  offset  to  a  6°  12'  curve 
from  a  point  24  feet  from  the  middle  of  a  chord 
104  feet  long. 


0 

2R 
R,  found  as  above,  =  925  feet. 

n  _  76  X  28  _  76  X  28  _ 
"  2~X92~5  "      1850 

Setting  for  0  is:  Opposite  76  on  the  D  scale 
set  185  on  the  C  scale;  shift  the  slide  to  the 
left  thru  its  entire  length  by  bringing  the 
runner  over  the  left  index  of  the  C  scale  and 
running  the  slide  thru  until  the  right  index  of 
the  C  scale  is  under  the  runner,  and  opposite 
28  on  the  C  scale  read  115  on  the  D  scale. 

ii.  Curves  are  usually  laid  out  by  deflection 
angles.  The  deflection  angle  is  the  angle 
between  a  tangent  at  a  given  point  and  a 
chord  thru  the  same  point.  The  deflection 
angles  ordinarily  used  for  laying  out  a  curve 
are  those  between  the  tangent  thru  the  be- 


SIMPLE   CURVES  17 

ginning  of  the  curve  and  the  chords  from  the 
beginning  of  the  curve  to  the  stations  on  the 
curve  and  to  the  end  of  the  curve.  / 

The  angle  between  two  chords  is  /measured 
by  one-half  of  the  arc  intercepted  by  them. 
If  this  arc  is  subtended  by  a  chord  100  feet 
long,  the  angle  between  the  two  chords  is 
J  D  (D  being  the  degree  of  the  curve).  If  the 
arc  is  subtended  by  any  other  length  of  chord, 
its  value  is  found  as  follows: 

Let  c  be  the  chord  subtending  the  arc  and 
d  be  the  angle  at  the  center  subtended  by 
the  arc.  Then  (as  already  shown  by  Fig.  2), 

c  =  2  R  sin  x  d   or   sin  ~  d  =  •=-„.    Sin  ~  D  = 

2 

100      ,-,        sin 

2R-     TheniiT 

and  D  are  small  and  for  small  angles  the  sines 

are  proportional  to  the  angles. 

sin  ^  d  _  d_        c  ,  _     c 

~      ~          (          " 


D~  100  "TOO 

The  application  of  this  is  shown  in  the  fol- 
lowing problem. 

Problem  9.  Find  the  deflections  for  a  6°  12' 
curve  with  a  central  angle  of  45°  36'  and  its 
P.O.  at  station  106  +  42.  The  length  of  any 


18 


THE    SLIDE    RULE 


curve  may  be  found  in  stations  by  dividing  the 
central  angle  by  the  degree  of  the  curve. 
Then 


L  = 


45°  36'      2736 


=  7.355  stations. 


6°  12'     ~  372 

P.O.  at   106  +  42 
L       =       7  +  35.5 
P.T.  at   113  +  77.5 

Let  d\  be  the  deflection  for  station  107  and 
z  the  deflection  for  the  last  chord. 


372  X  58 
~200~ 


_ 


(58  feet  from  106  +  42  to  107). 
_  372  X  77.5  _ 
~~ 


(77.5  feet  from  113  to  P.T.). 


Station. 

Deflection. 

106  +  42  P.O. 

107 

1°  48' 

108 

4°  54' 

109 

8°  00' 

110 

11°  06' 

111 

14°  12' 

112 

17°  18' 

113 

20°  24' 

113  +  77.  5  P.T. 

22°  48' 

SIMPLE   CURVES  19 

Setting  for  di  and  d2  is:  Opposite  372  of 
the  D  scale  set  2  on  the  C  scale  and  opposite 
58  and  77.5  on  the  C  scale  read  108  and  144 
respectively  on  the  D  scale. 

The  deflection  for  station  107  is  108'  =  1°  48'. 
The  deflection  for  station  108  is  found  by  add- 
ing 3°  06'  (|  D°)  to  the  deflection  for  station 
107.  The  deflection  for  station  109  is  found 
by  adding  3°  06'  to  that  for  station  108,  and  so 
on,  until  the  deflection  for  station  113  is  found. 
The  deflection  for  station  113  +  77.5  is  found 
by  adding  144'  =  2°  24'  to  the  deflection  for 
station  113.  As  the  deflection  for  station 
113  +  77.5  is  one-half  of  the  central  angle 
of  the  curve,  the  computation  checks. 

12.  Another  method  of  locating  a  curve  is 
by  offsets  from  the  tangent  thru  any  given 
point  of  the  curve.  The  method  is  shown  in 
Problem  10. 

Problem  10.  Find  the  necessary  distances 
to  locate  a  6°  12'  curve  from  station  91  +  40 
to  station  94,  by  offsets  from  the  tangent  thru 
station  91  +  40. 

60  6°  12'      60  X  186'  , 


=  1°  51.5'. 
<  C'  92  93  =  2  X  1°  51.5'  +  9-^  =  6°  49'. 


20 


THE    SLIDE   RULE 


<  D'  93  94  =  6°  49'  +  6°  12'  =  13°  1'. 

B  92  =  60  sin  1°  51.5'  =    1.86  feet. 
C"  93  =  100  sin  6°  49'  =  11.85  feet. 
Dr  94  =  100  sin  13°  1'   =  22.50  feet. 
C93  =  £  92  +  C'  93  =  13.71  feet. 
Z)'94  =  36.21  feet. 


91+40 


In  Fig.  7, 


A  92  -  AB  = 


Fig.  7 


£92 


1.862 


=  0.03, 


or 


2  X  A  92       120 
AB  =  60  -  0.03  =  59.97  feet. 


92 C"  =  9293  -     °  _-  =  100  -  ^^ 
2  X  92  93  200 

=  100  -  0.7,  or  EC  =  99.30  feet. 


931X  =  93  94  -         —  —  =  100  - 

2  X  93  94  200 

=  100  -  2.53,  or  CD  =  97.47  feet. 


SIMPLE   CURVES 


21 


AC  = 

AD  = 


159.27  feet. 
256.74  feet. 


The  settings  are  similar  to  those  given  in 
Problems  5  and  7. 

13.  It  is  sometimes  desired  to  change  a 
curve  so  that  it  may  end  in  a  parallel  tangent 
nearer  to  or  further  from  the  center  than  the 
tangent  in  which  it  ends.  Three  cases  arise 
in  practice,  as  follows: 


C" 


Fig.  8 


1st.  By  using  same  degree  of  curve,  find 
new  P.O.  and  P.T. 


AA'  =  VV  =  BE'  =  - 


-f- 
sm/ 


22 


THE    SLIDE    RULE 


Add  AAf  to  station  of  old  P.O.  for  the  sta- 
tion of  the  new  P.O.  and  add  AA'  to  station 
of  the  old  P.T.  for  the  station  of  the  new  P.T. 

2nd.  By  keeping  same  P.O.,  find  new  de- 
gree of  curve  and  new  P.T. 


In  Fig.  9, 


Fig.  9 


VV  -  - 

V     V  • 


-- 

•         -r 

sin  I 
AV  =  AV  +  VV. 

AV  = 


AV  =  R' 


SIMPLE   CURVES 


23 


2 
R'  = 


tan  ~  sin  / 


3rd.  By  ending  the  curve  directly  opposite 
old  P.T.,  find  new  degree  of  curve,  new  P.O. 
and  new  P.T. 


F     F' 


In  Fig.  10, 


VB  =  V'B'  +  VF. 
VB  =  R  tan  |  7. 
V'B'  =  Rr  tan  \  I. 

VF  =  ^~- 
tan/ 


24  THE   SLIDE   RULE 

Then 


(11) 


rv         r> 
it    =  It  — 


P 


tan  J  /  tan  / 

=  (£-#')  tan  7. 


From  Equation  (11), 
(R  -fi')tan7  =: 


tan^7 

Then  A  A'  =        ^  y- 

tanf  / 


Add  to  the  station  of  the  old  P.O.  AA'  for 
the  station  of  the  new  P.O.  From  Rf  the 
new  degree  of  the  curve  can  be  found.  From 
the  new  degree  of  the  curve  and  the  central 
angle  the  length  of  the  new  curve  can  be 
found,  and  from  the  new  station  of  the  P.O. 
and  the  length  of  the  new  curve  the  new 
station  of  the  P.T.  can  be  found. 

These  equations  are  better  adapted  to  the 
slide-rule  solutions  than  the  ordinary  ones  for 
these  problems.  If  the  new  ending  tangent 
is  nearer  the  center,  a  change  in  the  signs  will 
result  in  the  above  equations. 

Problem  11.  Find  the  stations  of  the  new 
P.O.  and  P.T.  for  ending  a  3°  30'  curve  in  a 


SIMPLE   CURVES  25 

parallel  tangent  13  feet  further  from  the 
center,  keeping  the  degree  of  the  curve  the 
same.  The  central  angle  is  22°  24'  and  old 
P.O.  is  at  station  126  +  40. 

13 


~       or»o  CkA 

sin  22  24 

Setting:  Opposite  13  on  the  A  scale  set 
22°  24'  on  the  S  scale  of  the  slide  reversed  and 
opposite  the  right  index  of  the  S  scale  read 
34.1  on  the  A  scale. 

22°  24'      1344 


Old  P.O.  at  station     126  +  40 

34.1 


New  P.O.  at  station  126  +  74.1 
L  =  6  +  40 


New  P.T.  at  station   133  +  14.1 

Problem  12.  Using  the  same  curve  as  in 
Problem  11,  it  is  desired  to  move  ending  tan- 
gent 13  feet  nearer  the  center,  keeping  the 
same  P.O. 


R'  =fl- 


P 


I  .    T 
tan  ~  sin  / 


26  THE   SLIDE   RULE 

p      100  X  3438  . 

R  = ^ph =  1"35  feet. 

ZiL\J 

P  =  13 

tan  i  /  sin  /      tan  11°  12'  sin  22°  24' 

#'  =  1635  -  172.1  =  1462.9  feet. 

13 

Setting  for  tan  11°  12'  sin  22°  24' is:  °PP°- 
site  13  on  the  A  scale  set  22°  24'  on  the  S  scale 
of  the  slide  reversed  and  opposite  the  right 
index  of  the  S  scale  read  341  on  the  A  scale; 
opposite  341  on  the  D  scale  set  11°  12'  on  the 
T  scale  and  opposite  the  left  index  of  the  T 
scale  read  172.1  on  the  D  scale. 

100  X  3438  _ 

1463 

22°  24' _  1344' _ 
=  WW~~"  234r  = 
P.O.  at  126  +  40 

P.T.  at  132  +  15 

Problem  13.  Using  the  same  curve  as  in 
No.  11  and  ending  new  curve  directly  opposite 
old  P.T.  in  a  new  parallel  tangent  13  feet 
further  from  the  center,  find  new  degree  of 
curve  and  new  stations  of  the  P.O.  and  the 
P.T. 


SIMPLE   CURVES  27 


R>  =  R  --  -     —  _  =  1635 
tan  J  /  tan  / 


10 

_  i  KQ 


___  _  _ 

tan  11°  12'  tan  22°  24' 

Rf  =  1476  feet. 
Setting  for  tanllol2^an22024/  is:   Oppo- 

site 13  on  the  D  scale  set  11°  12'  on  the  T 
scale  of  the  slide  reversed,  bring  the  runner 
to  the  right  index  of  the  T  scale,  bring  22°  24' 
on  the  T  scale  under  the  runner  and  opposite 
the  left  index  of  the  T  scale  read  159  on  the 

D  scale. 

zy_  100X3438 

1476 

.,      22°  24'      1344 
L  =  - 


AA'  =  (B- 


=  159  tan  11°  12' + 


sin  22°  24' 
=  31.5 +  34.1  =  65.6  feet. 
Old  P.O.  at  126  +  40 

65.6 

New  P.O.  at  127+   5.6 

L'  =  5  +  78 

New  P.T.  at  132  +  83.6 


28 


THE   SLIDE   RULE 


14.  Other  special  problems  in  simple  curves 
may  be  solved  with  fair  precision  by  the  use 
of  the  slide  rule.  To  get  good  results  it  is 
necessary  that  the  angles  whose  sines  are  used 
shall  be  between  0°  and  70°,  and  those  whose 
cosines  are  used  shall  be  between  20°  and  90°. 
The  following  is  a  sample  of  such  special 
problems. 

Problem  14.  Find  the  degree  of  the  simple 
curve  that  will  pass  thru  a  given  point  and 
will  join  two  tangents  having  a  given  inter- 
section angle.  The  values  are  given  in  Fig.  11. 

A  16*30'      V 


In  Fig.  11, 


Fig.  11 


=  90°  -  34°  45'  =  55°  15 


OV= 


SIMPLE   CUEVES  29 

R. 
R 


cos  18°  15' 
In  the  triangle  07P, 

R  sin  55°  15' 


R  sin  OPV 


cos  18°  15' 

.          v      sin  55°  15'  _  sin  55°  15' 
cos  18°  15'       sin  72°  45'' 

Solving  this  by  the  slide  rule, 
sin  OPV  =  sin  60°. 

The  angle  OPV  is  greater  than  90°,  as  seen 
by  the  construction  of  Fig.  11.     Hence 
OPV  =  180°  -  60°  =  120°. 

The  setting  of  the  slide  rule  for  finding  the 
sin  OPV  is  as  follows: 

Reverse  the  slide  in  the  rule.  Make  the 
indices  of  the  A  scale  and  the  sine  scale  to 
coincide,  move  the  runner  over  55°  15'  on 
the  sine  scale,  bring  71°  45'  on  the  sine  scale 
under  the  runner,  move  the  runner  to  the 
right  index  of  the  sine  scale  and  then  set  the 
indices  of  the  sine  scale  to  agree  with  the 
indices  of  the  A  scale,  reading  the  result  60° 
under  the  runner. 


30  THE   SLIDE   RULE 

POV  =  180°  -  (120°  +  55°  15')  =  4°  45'. 
sin  55°  15' 


—  —  A0   .    t  • 

sm  4  45 

This  solved  by  the  slide  rule  gives 
R  =  1190  feet. 


D,  in  minutes,  =       -  =  289'  =  4°  49'. 

ri 

It  is  probable  that  a  5°  curve  would  be  used 
under  the  conditions  given. 

15.  In  the  remaining  problems  the  settings 
of  the  slide  rule  will  not  be  given,  except  in 
a  few  cases,  as  those  already  given  show  the 
methods  that  usually  may  be  applied. 


CHAPTER  III 
COMPOUND  CURVES 

1 6.  It  is  sometimes  necessary,  for  the  best 
location,  to  combine  curves  of  different  radii. 
When  these  curves  are  on  the  same  side  of 
their  common  tangent,  they  form  a  compound 

A F         V 


Fig.  12 


curve  and  when  on  opposite  sides  of  their 
common  tangent,  they  form  a  reversed  curve. 
A  reversed  curve  may  have  branches  of  equal 
or  unequal  radii.  In  a  compound  curve  the 
radii  of  the  branches  are  always  unequal. 
31 


32  THE   SLIDE   RULE 

Fig.  12  shows  a  compound  curve,  the  tangent 
distances,  TI  and  T8,  for  which  may  be  found 
as  follows: 

Ti  =  AV.  Ts  =  VB. 

Ri  =  AO.  Rs  =  CB. 

FE  =  FN  +  NE  =  Ri  tan  J  Ii  +  Rs  tan  J  I8. 
FV  and  VE  may  be  found  by  solving  the 
triangle  FVE. 

Then 

Tl  =  Ei  tan  £  7,  +  **7. 
T8  =  £8  tan  J  7a  +  7#. 

Problem  15.  Find  the  tangent  distances  for 
the  compound  curve  consisting  of  a  3°  curve 
with  a  central  angle  of  14°  20'  beginning  at  A 
and  compounding  with  a  6°  curve  with  a 
central  angle  of  40°  40'  ending  at  B. 

p       100  X  3438  f 

=  =  191°  feet- 


100  X  3438      n_  f 

=955feet' 


-860 

=  1910  tan  7°  10'  +  955  tan  20°  20' 
=  240  +  354  =  594. 


F  V  =  o  sin  40°  40'  =  473. 

sin  55° 


COMPOUND   CURVES  33 


VE  =    .     *    sin  14°  20'  =  179.5. 
sin  55 

r,  =  240  +  473  =  713  feet. 
T8  =  354  +  179.5  =  533.5  feet. 

17.  Nearly  all  problems  in  compound  curves 
may  be  solved  by  the  method  of  coordinates, 
the  application  of  which  is  given  in  the  follow- 
ing three  problems. 

Problem  16.  The  deflection  angle  of  the 
tangents  of  a  compound  curve  is  52°.  The 
curve  begins  at  station  372  and  the  first  branch 
is  a  6°  curve  to  the  left,  5  stations  long.  The 
T8  is  to  be  550  feet.  Find  the  radius  and  the 
degree  of  the  second  branch  and  the  stations 
of  the  P.C.C.  and  the  P.T. 

In  Fig.  12,  the  P.O.  of  the  curve,  to  fulfil 
the  condition  given  in  the  problem,  is  at  B. 

I8  =  6  X  5  =  30°. 

Ii  =  I  -  I3  =  52°  -  30°  =  22°. 

=  955'. 


The  lengths  and  directions  of  the  lines  NC, 
CB  and  BV  are  known,  and,  using  NC  as  a 
meridian,  their  Z/'s  and  M's  are  found  as 
follows: 


34 


THE   SLIDE   RULE 


Azimuth 

L 

M 

+ 

- 

•f 

- 

NC 

0°00' 

Feet. 
955 

955 

CB... 

210°  00' 

955 

827 

477 

BV 

120°  00' 

550 

275 

476 

VN  

955 
147 

1102 

476 
1 

477 

F V 


Fig.  12 

k  cos  210°  =  -  955  cos  30°  =  -  827. 

(cos  30°  =  sin  60°.) 
12  sin  210°  =  -  955  sin  30°  =  -  477. 
k  cos  120°  =  550  cos  60°  =  275. 

(cos  60°  =  sin  30°.) 


COMPOUND   CURVES  35 

Z3  sin  120°  =  550  sin  60°  =  476. 
Tan  of  azimuth  of  VN  =  TiT  =  0.0068, 
or  azimuth  of  VN  =  23'. 

Length  of  VN  =  147  +  2  ^^  =  147.03, 

practically  147  feet. 

In  Fig.  12,  the  azimuth  of  VA  is   120°  - 
52°  =  68°.     The  azimuth  of  VN  is  0°  23'  and 
of  NA  is  the  angle  ANO  =  \  (180°  -  22°)  = 


Fig.  13 

79°.     To  find  the  lines  AV  and  NA,  the  tri- 
angle shown  in  Fig.  13  is  solved: 


AV  =    ,     'sin  101°  23'  =  755  feet. 


147 
sin  11° 

147 
sin  11 

NA  711 


67°  37' =  711  feet. 

=  1865  feet. 


\Ii      2  sin 

X  100  =  184.5'  =  3( 


36 


THE   SLIDE   RULE 


In  practice  a  3°  curve  would  probably  be  used. 

22° 
Li  =  -30  =  7  +  33.3 

P.C.C.  is  at         377 


P.T.  is  at  384  +  33.3 

Tl  =  AV  =  755  feet. 

A  755'  V 


Fig.  14 


Problem  17.  Given  the  data  shown  in  Fig. 
14,  find  the  degree  of  the  second  branch  of  the 
compound  curve  and  the  stations  of  the  P.C.C. 
and  the  P.T.,  beginning  the  compound  curve 
with  a  3°  (the  long  radius  branch)  curve  at 
station  78. 

Use  OA  as  a  meridian. 


COMPOUND  CURVES 


37 


T  in/» 

T  AncrtVi 

L 

M 

+ 

- 

+ 

-. 

OA. 

(TOO' 

Feet. 
1910 

1910 

AV 

90°  00' 

755 

755 

VB 

142°  00' 

550 

434 

338 

BO 

1910 

434 
1476 

1093 

1693 

L3  =  550  cos  142°  =  -  550  sin  52°  =  -  434. 
M  3  =  550  sin  142°  =  550  sin  38°  =  338. 

For  the  line  OB  the  coordinates  would  be 
positive. 


Tan  of  the  azimuth  of  OB  = 


1093 


=  0.741. 


1476 

Az.  of  OB  =  36°  30'. 
Az.  of  BO  =  216°  30'. 
Az.  of  EC  =  142°  +  90°  =  232°. 

Angle  a  =  232°  -  216°  30'  =  15°  30'. 

In  the  triangle  CBO,  EC  =R8y  CO  =  Rt  -  R9, 


OB  =   . 


sin  36°  30' 


=  1838  feet. 


s  = 


^=919+^2=1874. 


38  THE    SLIDE    RULE 

i      =  (s  -  ft)  (s  -  1838)      (1874  -  R.)  36 

1838ft  1838ft 

=  1874  -  ft 

51ft 
Sin2  J  (15°  30')  =  sin2  7°  45'  =  0.01815. 

The  setting  for  sin2  7°  45'  is  as  follows: 
Reverse  the  slide  in  the  rule,  set  the  indices  of 
the  sine  scale  and  the  A  scale  coincident  and 
read  on  the  A  scale  1348  opposite  7°  45'  on  the 
sine  scale.  Then  set  the  runner  over  1348  on 
the  D  scale  and  read  1815  on  the  A  scale  under 
the  runner. 

0.01815  X  51  X  ft  =  1874  -  ft, 
or  ft  =  974  feet. 

To  find  51  X  0.01815,  keep  the  runner  in  the 
position  stated  above,  bring  the  numbered  face 
of  the  slide  to  the  front,  set  the  left  index  of 
the  B  scale  under  the  runner  and  read  925  on 
the  A  scale  opposite  51  on  the  B  scale. 


Use  a  6°  curve. 

Sinl5°30' 


Ia  =  31°  40'. 


COMPOUND  CURVES  39 

It  =  52°  -31°  40'  =  20°  20'. 
Lj- 

L,= 

U 

P.C.         at 

P.C.C.     at 

L8 

P.T. 


=  5  +  27.7  stations. 

86  +  00 

6  +  77.8 
92  +  77.8 

5  +  27.7 


at      98  +  05.5 
Problem  18.   Begin  a  compound  curve  with 
a  4°  (long  radius)  curve  at  station  378.     Find 
the  degree  of  the  second  branch  and  stations  of 


878 


0 


Fig.  15 


P.C.C.  and  P.T.  of  the  compound  curve  to 
connect  A  and  B  under  the  conditions  given 
in  Fig.  15. 


THE   SLIDE    RULE 


Length 

L 

If 

+ 

- 

+ 

- 

OA 

0°00' 

Feet 
1432  5 

1432  5 

AB 

116°  00' 

1200 

525 

1079 

BO  ... 

907.5 

1079 

or 


Using  OA  as  a  meridian. 

1200  cos  116°  =  -  525. 
1200  sin  116°  =  1079. 
1079 

907.5'      . 
azimuth  OB  =  49°  56'. 
Azimuth  BC  =  240°  00' 
Azimuth  BO  =  229°  56' 


Tan  azimuth  OB  = 


a  =    10°  04' 
1079 


B0  = 


r,  =  1408.* 


sin  49°  56' 

s  =  |  (1432.5  -  Ra  +  Rs  +  1408)  =  1420.25. 
gin*1^  (s~Rs)(s-BO) 

RaW 

(1420.25  -  R.)  12.25 
1408  E, 

*  To  get  a  good  result  care  must  be  taken  to  read 
this  value  on  the  slide  rule  very  carefully. 


COMPOUND  CURVES  41 

(1*20-25  -ft)  12-2) 


00077 

1408  fl, 

or  R3  =  755  feet. 

3438  X  100 


Use  a  7°  30'  curve. 
l^Rs  =  1432.5  -  755  =  677.5. 


Sin  COB  =  j££=  sin  10°  04', 
677.5 

or          COB  =  11°  12'. 

78  =  10°  04'  +  11°  12'  =  21°  16'. 
/,  =  60°  -  21°  16'  =  38°  44'. 

3«°  44' 
£i  =        F-  =  9  +  68.3. 


21°  16' 


•L<«           yo 

P.O. 

P.C.C. 
L, 
P.T. 

30' 

at 

at 
at 

-    A  T  OO.il. 

378 
9  +  68.3 

387  +  68.3 
2  +  83.5 

390  +  51.8 

Problem  19.  A  4°  curve  begins  at  station 
395  +  30  and  ends  at  station  408  +  40.  Find 
the  station  of  the  P.C.C.,  where  a  6°  curve 


42 


THE   SLIDE   RULE 


will  compound  so  as  to  end  in  a  parallel  tangent 
12  feet  nearer  the  center. 


r>  T>  77'D 

=  KI   —  Ks   =  xI/jD. 

BE'  =  (Rt  -  Rs)  vers.     I8 

=  (Ri  —  Ra)(l  —  cos  I8). 
Ri  -  Rs  =  1432.5  -  955  =  477.5. 

12     =0.9749. 


Fig.  16 

Sin  (90  -  7.)  =  0.9749,   or  90  -  Is  =  77°.* 
L  =  13°. 

7  =  13.10  X  4  =  52°  24'. 
Il  =  52°  24'  -  13°  =  39°  24'. 

*  Reading  for  this  must  be  carefully  made. 


COMPOUND  CURVES  43 

100 

Ls  =  ^  =  2  +  16.7. 
Li  =  39^24'  =  g  +  85. 
P.O.       at      395  +  30 


P.C.C.   at      405  +  15 
L3  2+  16.7 

P.T.       at      407  +  31.7 

• 

Problem  20.  An  8°  curve  begins  at  station 
372  and  ends  at  station  378  +  50.  Find  the 
stations  of  the  P.C.C.'s  and  the  P.T.  for  a 
three-centered  compound  curve,  with  one  sta- 
tion of  a  1°  curve  at  each  end  and  passing  thru 
the  same  P.C.  and  P.T.  as  the  8°  curve. 

sin  ^7       Ri  -  Rs 


Oi03      sin  i  /*"#*-#' 
/  =  6.5  X  8  =  52°. 
7S  =  52°  -  2°  =  50°. 
RL-  R  =  5729.7  -  716.8  =  5012.9. 


Rl-R8=  5012.9  =  5200. 

Rs  =  5729.7  -  5200  =  529.7  feet. 


44 


THE   SLIDE   RULE 


L  -    5°°     -4  +  61 
**         ^'  ~ 


P.O. 
LL 
P.C.C.i 
L8 

P.C.C.2 

Lt 
P.T. 

Fig. 
at 

at 
at 
at 

17 

372 
1 

+  00 
+  00 

373 

4 

+  00 
+  61 

377 
1 

+  61 
+  00 

378 

+  61 

COMPOUND  CURVES 


45 


Problem  21.  Given  two  straight  parallel 
tracks  with  center  lines  13  feet  apart  and  with 
a  long  chord  of  186.4  feet,  find  the  degree  of 
the  equal  branches  of  a  reversed  curve  to  con- 


'0 


Fig.  18 


nect  the  parallel  tracks.    A  ED  and  AOD  are 
similar  triangles,  then 


46 


THE    SLIDE    RULE 


13        4  X  13 
Rr  =  667  feet. 
3438  X  100 


=  667. 


667 


=  515'  =  8°  35'. 


Problem  22.  Given  the  conditions  shown  in 
Fig.  19,  find  the  degree  of  the  reversed  curve, 
with  branches  of  equal  radii,  connecting  the 


Fig.  19 

tangents  AB  and  CD,  and  having  EC  as  the 
common  tangent  of  the  two  branches,  and  also 
find  the  stations  of  the  P.C.,  P.R.C.  and  P.T. 

940  =  Rr  tan  18°  +  Rr  tan  14°. 
940 


tan  18°  +  tan  14' 
940 


0.325  +  0.2495 


=  1640  feet 


COMPOUND  CURVES  47 

_  3438  X  100 

1640 

Use  3°  30'. 

R  for  3°  30'  curve  =  1637  feet. 
Tl  =  1637  tan  18°  =  532  feet. 
376  +  00 
5  +  32 


36 


P.O.      at    370  +  68      Lx  =  ~^  =  10  +  28.6. 
10  +  28.6 


P.R.C.  at   380  +  96.6  L2  =  f-  =  8  +  00. 

o.O 

8  +  00.0 
P.T.      at    388  +  96.6 


CHAPTER  IV 

THE  VERTICAL  CURVE 

18.  The  vertical  curves,  used  at  sags  and 
summits  to  connect  adjacent  grades,  or  used 
elsewhere  to  connect  grades  of  the  same  kind, 
are  usually  parabolas.  The  following  problem 
shows  how  the  elevations  on  such  a  curve  may 
be  found. 

Problem  23.  Find  the  elevations  of  the  sta- 
tions on  a  ten-station  vertical  curve  to  connect 
the  grades  shown  in  Fig.  20. 


Fig.  20 

The  offsets  from  the  tangent  to  the  parabola 
vary  as  the  squares  of  the  distances  from  the 
beginning  of  the  curve  to  the  points  where  the 
offsets  are  made.  The  beginning  of  the  ver- 
tical curve  is  at  station  225  and  the  end  is  at 


THE   VERTICAL    CURVE 


49 


station  235.  The  elevation  of  station  225  is 
515  and  of  station  235  is  517.5.  The  vertical 
offset  from  a  point  on  the  prolongation  of  the 
1-per  cent  grade  to  station  235  is  7.5  feet.  By 
the  above  rule  the  offset  at  station  226  is  T^ 
of  7.5  or  0.075.  The  other  offsets  may  be 
found  from  this  by  the  above  rule. 


Station. 

Elevation  on 
the  straight 
grade. 

Offset  . 

Elevation  on 
the  curve. 

225 

515.00 

515  000 

226 

516.00 

0.075 

515.925 

227 

517.00 

0.300 

516.700 

228 

518.00 

0.675 

517.325 

229 

519.00 

1.200 

517.800 

230 

520.00 

1.875 

518.125 

231 

521.00 

2.700 

518.300 

232 

522.00 

3.675 

518.325 

233 

523.00 

4.800 

518.200 

234 

524.00 

6.075 

517.925 

235 

525.00 

7.500 

517.500 

The  setting  for  the  offsets  is  as  follows: 
Set  the  right  index  of  the  slide  opposite  75  on 
the  right-hand  A  scale,  and  opposite  2,  3,  4, 
etc.,  on  the  C  scale  read  3,  675,  12,  etc.,  on  the 
A  scale.  By  rough  figuring  obtain  the  posi- 
tions of  the  decimal  points. 

A  check  on  the  elevation  of  station  230  on 
the  curve  may  be  found  from  the  principle 


50  THE   SLIDE    RULE 

that  the  parabola  is  midway  between  its  chord 
and  its  vertex  at  its  middle  point.  The  eleva- 
tion of  the  chord  at  its  middle  point  is  the 
mean  of  the  end  elevations  The  elevation  of 
the  middle  of  the  curve  is  the  mean  of  the 
elevation  of  the  middle  of  the  chord  and  the 
elevation  of  the  vertex.  Applying  this  to  the 
above  problem,  the  elevation  of  the  middle  of 

the  chord  is  -     — ^ —    -  =  516.25  and  the  ele- 

.  516.25+520 
vat  ion  of  the  middle  of  the  curve  is  —    — - — 

£ 

=  518.125,  which  checks  the  elevation  found. 

By  the  extension  of  this  principle,  the  eleva- 
tions of  points  on  the  vertical  curve  may  be 
found  or  a  parabola  may  be  laid  out. 


CHAPTER  V 
TURNOUTS 

19.  Fig.  21  is  a  line  diagram  representing 
a  split  switch  turnout,  the  one  now  most 
commonly  used.  AH  and  LK  are  the  movable 


A 


* 


Fig.  21 

rails.  The  rails  A G  and  BP  are  continuous. 
F  is  the  point  of  the  frog.  The  rails  HF  and 
QP  are  curved  and  the  rail  AH,  when  the 
turnout  is  set  for  the  branch  or  siding,  is  tan- 
51 


52  THE   SLIDE    RULE 

gent  to  the  rail  HF.     AH  is  the  switch  rail, 
I,  and  HE  is  equal  to  the  amount  of  its  throw,  t. 

EAHj  the  switch  angle,  =  8. 
a.     e      EH      t 

=  AS==r 

AB,  the  gage  of  the  track,  =  g. 
•  The  standard  gage  is  4'  8J". 
FOD  =  F,  the  frog  angle. 
0V  —  Rj  the  radius  of  the  turnout  curve. 
(R  +  iflf)  cos  AS  -  (R  +  ±g)cosF  =  g  -t. 

R  +  ^g==  cosS-cosF' 

or  #  = 1 ^  —  \  g. 

cos  £  —  cos  F 

The  distance  BF  is  called  the  lead,  L. 

KF. 


Angle    #/^C  =  F  -  f  F0#  =  F  -  J  (F- 


r_7 


tan 


1        =  2(R+ig)sm%(F-S)  .    (13) 


TURNOUTS 


53 


Fig.  22  shows  the  form  of  the  frog.     The 

,  ,,     ,       .    FR      ...     FN  . 
number  of  the  frog  is  -= ,  altho          is  some- 
times used. 


Fig.  22 
Where  n  is  the  number  of  the  frog, 

f  =  A.     . 


(15) 


Problem  24.  Find  the  radius  and  the  lead 
of  a  split  switch  turnout  using  a  No.  9  frog, 
18-foot  switch  rail,  a  S^-inch  throw  and 
standard  gage  track. 


o-     a  _  *  _ 

* 


5'5 


12X8 

S  =  1°  28'. 


5.5 
216 


=  0.0254. 


9^  =  «VQ  =  ili  =  0-0556. 

A  71          £  /\  y          J.O 

For   small   angles   sines   and  tangents  are 
practically  equal,  and  for  all  angles  less  than 


54  THE    SLIDE    RULE 

5°  45'  the  sine  in  place  of  the  tangent  scale  of 
the  slide  rule  should  be  used  for  tangents  of 
such  angles. 

By  use  of  the  S  scale  of  the  slide,  -|  F  = 
3°  11'  or  F  =  6°  22'. 

Setting  for  these  is:  Opposite  right  and 
left  indices  of  the  A  scale  set  the  indices  of 
the  slide  reversed  and  opposite  0.0254  on  the 
A  scale  read  1°  27'  on  the  S  scale,  also  op- 
posite 0.0556  on  the  A  scale  read  3°  11'  on 
the  S  scale. 


r  tan  $(F  +  8)  12  tan  3°  55' 

L  =  18  +  62.2  =  80.2  feet. 


R  +  ^  ~  24  sin  3°  5?'  sin  2°  27'  =  ?28 
72  =  725.65  feet. 


20.  Where  the  turnout  leaves  the  main 
track  on  a  curve,  the  radius  and  the  lead  may 
be  found  by  the  precise  or  approximate  method 
following. 


TURNOUTS 


55 


Precise  method: 

1st.  When  the  turnout  is  on  the  inside  of 
the  main  track  curve.  In  Fig.  23,  0  is  the 
center  of  the  main  curve,  C  is  the  center  of  the 
turnout  curve,  F  is  the  frog  point  and  AB  is 
the  gage  of  the  track. 


Fig.  23 
In  the  triangle  AFO, 


2' 

"2/_    g 


Rm 


56  THE    SLIDE    RULE 

tan^  ff  =      g 
cot  |  0      2  Rm' 
i  ^      9  cot  i  F 

= 


taniO  =  f^  ......     (16) 

Km 

In  the  triangle  BCF,  BC  =  Rt  -  | 

z 

and  F(7  =  ^  +    , 


nr  r,  .         g  cot  j  F        _  g^  ,    . 

~  ~  '   l    ; 


2nd.  When  the  turnout  is  on  the  outside  of 
the  main  track  curve.  In  Fig.  24,  solving  the 
triangle  AFO  by  the  method  used  above, 


Then  solving  the  triangle  BCF, 


tan  i  (F  -  0) 


TURNOUTS 


57 


In  either  case,  Lt  =  2  Rm  sin  \  0, 

or  Lt  =  2  Rt  sin  %  (F  d=  0). 


Fig.  24 

21.  Fig.  25  shows  a  turnout  from  a  straight 
track  where  a  stub  switch  is  used.  BF  is  the 
lead  and  AB  is  the  gage. 

Ls  =  BF  =  AB  cot  i  F  =  2  0n.    .     (18) 


58 


THE    SLIDE   RULE 


Make  BC  =  g.    Then  the  triangles  AFC  and 
AOF  are  similar  and  AO:AF  ::AF:AC 
ACXAO      =  I/. 

2g(R,+g^='A~f,     or    2  gR,  =  Af-g\ 

?8  =  20n2.  (19) 
(20) 


or 
gn 


Fig.  25 

22.   The  approximate  method  for  turnout 
from  curved  main  track. 

Let    R8  =  radius  of  turnout  from  a  straight 

track  using  a  given  number  of 

frog. 


TURNOUTS  59 

Rt  =  radius    of   turnout   from    curved 
track  using  same  number  of  frog. 
Rm  =  radius  of  curved  main  track. 

From  Equations  (16),  (17)  and  (20),  for  the 
condition  where  the  turnout  is  on  the  inside  of 
the  curved  main  track, 

p  .  P    .  p  .       9n  9n  9n 

'•tan  %F  '  tan  J  0  'tan  \  (F  +  0) 

As  F  and  0  are  small  angles,  the  tangents  may 
be  taken  equal  to  the  arcs  of  unit  radius,  or 


••jp-O'F+0 

The  degrees  of  curves  are  practically  inversely 
as  their  radii,  or 

D8:Dm:Dt::F  :  0  :  F  +  0. 

By  composition  (Ds  +  Dm)  :  Dt  : :  F  +  0 :  F  +  0. 
Hence  Dt  =  D8  +  Dm. 

Where  the  turnout  is  on  the  outside  of  the 
curved  main  track,  a  similar  proportion  applies. 

Ds:Dm:Dt::F:0:F  -0. 
By  division 

D8-Dm:Dt::F-0:F-0. 
Hence  Dt  =  D3  —  Dm. 


60 


THE    SLIDE    RULE 


The  condition  of  Dt  =  Dm  —  D8  may  arise  as 
shown  in  Fig.  26,  where  0  is  the  center  of 
the  main  track  curve  arid  C  is  the  center  of  the 
turnout  curve. 

A 
B 


Fig.  26 

In  either  case,  Lt  =  2  Rt  sin  \  (F  =b  0),  and  as 
F  and  0  are  small  angles,  the  tangents  and 
sines  are  about  equal  or 

Lt  =  2  Rt  tan  \  (F  ±  0), 
gn 


Rt 


Hence 


=  2  gn. 


Problem  25.  Find  the  radius  and  the  lead 
of  a  turnout  on  the  inside  of  a  4°  main  track 
of  standard  gage,  using  a  No.  9  frog.  Both 


TURNOUTS  61 

the  precise  and  approximate  methods  of  solu- 
tion follow. 
Precise  method: 


4.708  X  9 


3438  X  100          ~ 
m  =         240 
tan  J  0  =  0.0296. 

Solving  by  the  slide  rule  using  the  sine  scale 
as  described  in  Problem  24,  \  0  =  1°  42'  and 
0  =  3°  24'.  F  =  6°  22'  as  in  Problem  24. 

4.709  X  9 

leet. 


>t  —  i  —  i  /ET  ,  —  Ty:      7  -  ,0  P0/ 
tan  i  (F  +  0)      tan  4°  53 

(Method  of  solution  as  given  in  Problem  24.) 
343000 


Lt  =  2Rtsm%  (F  +  0)  =  2  X  496  X  sin  4°  53' 
=  84.6  feet. 

Approximate  method: 
Dt  =  Dm  +  Ds. 

5730          2865  ooo 


4.708  X  92 


62  THE    SLIDE    RULE 

An  =       4°      0' 


A  =11°  29' 

L,  =  20n  =  2  X  4.708  X  9  =  84.6  feet. 


The  use  of  the  slide  rule  for  the  solution  has 
been  given  for  similar  formulas. 

23.  Fig.  27  shows  in  outline  the  connection 
between  a  parallel  siding  and  the  main  track 
by  a  turnout  and  a  connecting  curve  to  which 
the  following  problem  applies. 


Fig.  27 

Problem  26.  Find  R^  D%  and  Z/2,  where  the 
distance  between  track  centers  is  13  feet  and 
a  No.  9  frog  is  used. 

FA  and  EB  may  be  considered  as  the  rails 
of  a  track.  Then  the  curve  EF  may  be  con- 
sidered as  the  curve  part  of  a  turnout.  The 
gage  for  this  is  p  —  g. 


TURNOUTS 


63 


Then        #2-f  =  (p-g)2n*. 

R2  -  6.5  =  (13  -  4.71)  2  X  92 

=  8.29  X  2  X  92  =  1340. 
R2  =  1346.5  feet. 
3438  X  100 


D2  = 


=  255'  =  4°  15'. 


1346.5 
F  for  a  No.  9  frog  may  be  found  from 

=        =       =  0.055. 


Using  tangent  =  sine, 

£  =  3°  11',    or    F  =  6°22'. 


L2  =  #2  -      sin  ^  =  1344.3  sin  6°  22' 

=  149  feet. 

24.  Fig.  28  shows  a  connection  between  two 
parallel  tracks  by  the  use  of  two  similar  turn- 


Fig.  28 

outs  and  a  piece  of  straight  track  whose  center 
line  is  a  common  tangent  to  the  center  lines 


64  THE   SLIDE   RULE 

of  the  curves  of  the  turnouts.  The  following 
problem  applies  to  this  form  of  a  connection. 
Problem  27.  Find  the  distances  a,  b  and  c 
for  the  connection  between  two  parallel  straight 
tracks  of  standard  gage,  with  13  feet  between 
track  centers  and  with  No.  9  frogs  used  in  the 
turnouts. 

F  =  6°  22'  (from  Problem  26). 

a  =  c  =  2  gn  =  9.42  X  9  =  84.7  feet. 

b  =  p-g-gsecF  =  8.292  -  4.71  sec  6°  22' 
tan  F  tan  F 

4'71 


82Q2- 


cos  6°  22'      8.292-4.73 


tan  F  tan  6°  22' 

=  31.9  feet. 

25.  Fig.  29  shows  the  connection  between 
the  parallel  siding,  on  the  outside  of  the  curved 
main  track,  and  the  main  track  by  a  turnout 
and  a  connecting  curve.  Solving  the  triangle 
OBG, 

7?     4-,    ^-9- 

*"  P      2 


—  j  —  i"?^  —  ^  T-V  —  ;  —  7: 

cot§0      2Rm  +  p  2 


„ 

A/w  +  p  —  2  +  ^  + 

.      1  0  _  (p  -  flf)  n 
or  oan  7:  L/  —  —————  . 


TURNOUTS 


Fig.  29 
Solving  the  triangle  PBH, 


Rt_+Rt_p+ 

z  * 


tan  |F  p  -  g 


66  THE   SLIDE   RULE 

r>  _  P  _  -n 

Kt         ~ 


Problem  28.  By  both  the  precise  and  ap- 
proximate methods  find  the  radius  and  lead  of 
the  turnout  and  the  radius  (R2)  and  length 
(L2)  of  the  connecting  curve  to  connect  a 
4°  30'  main  track  of  standard  gage  with  a 
parallel  siding  using  a  No.  9  frog.  The  siding 
is  on  the  outside  of  the  main  track,  with  13 
feet  between  the  track  centers. 

Precise  method: 

n      gn       4.708  X  9 
"" 


p        3438X100      107n, 
--  270~~     = 

Solving  by  the  slide  rule  using  the  sine  scale, 

J  0  =  1°  54'  and  0  =  3°  48' 
F  =  6°  22' 


F  -  0  =  2°  34' 
i  (F  -  0)  =  1°  17' 
4.708X9 

==  189°  feet> 


3438X100 
A=      ~1890~ 
Dt  =  3°  02', 


TURNOUTS  67 


_  100  X  3°  48'  _  100  X  228  _         , 
4°  30'  270 

tan  J0'  =  l£^  =  *§^  =  0.0584. 
P        p         1276.5 


Solving  by  the  slide  rule  using  the  sine  scale, 
1  0  =    3°  20' 
0=6°  40' 
F  =    6°  22' 
F  +  0  =  13°   2' 
i(F  +  0)=    6°  31' 
p  _       (p  -  g)  n       .  8.29  X  9  _ 
~  ~°'~ 


and 

#2  =  659.5  feet. 

_  3438  X  100  _    22  _  R0      , 
659.5  42' 

100  X  13°  02'      100  X  782          .  , 


Approximate  method: 

By  Problem  25,  D8  =  7°  29' 
Dm  =  4°  30' 


Dt  =  2°  59' 
=  84.6  feet. 


68  THE    SLIDE    RULE 

Let  Rs  =  radius  of  the  connecting  curve  if 
the  tracks  were  straight. 

fia  ~  |  =  2  (p  -g)  n*  =  2  X  8.29  X  92  =  1340  feet. 

Rz  =  1346.5  feet 

and 

3438  X  100 

1346.5 
Z>3  =  4°  15' 
Dm  =  4°  30' 


=  8°  45r 
3438  X  100 


L2  =  2(p-g)n  =  149  feet. 

Problem  29.  By  the  precise  and  approxi- 
mate methods  find  the  radius  and  the  length 
of  the  connecting  curve  for  a  parallel  siding  on 
the  inside  of  a  4°  main  track  of  standard  gage, 
with  13  feet  between  the  track  centers  and  using 
a  No.  9  frog. 

Precise  method: 


TURNOUTS  69 

Using  the  sine  for  the  tangent 

i  0  =  3°  00',  or  0  =  6°  00'. 
F  as  found  in  previous  problem  =  6°  22', 

F  -  0  =  22'. 

_p_      (p-g)n          8.29  X  9 
2      2~tani(^-0)      tan  0°  11'' 
tan  11'  =  approximately  11  X  0.00029 
=  0.00319. 


p  _  8.29  X  9 
"  2  ~  0.00319       23'39°' 


or 

#2  =  23,384  feet. 

3438  X  100       ,  c,  , 
2  =      23384        =        (approximately). 

99' 
£,_£!_  147  feet. 

1Q 

Approximate  method: 
Rs-£  =  2(p-  g)n*  =  2  X  8.29  X  92  -  1340  feet. 

Zt 

Rs  =  1346.5  feet. 

_  3438  X  100  _        ,  _    0      , 

1346.5 
D2  =  4°  15'  -  4°  =  15'. 

L2  =  2(p-g)n=  149  feet. 


70  THE    SLIDE    RULE 

26.  "Y"  curves  are  used  to  connect  the 
main  tracks  and  branch  tracks,  so  that  the 
trains  coming  in  either  direction  may  go  from 
the  main  tracks  to  the  branch  tracks  without 
backing.  A  "Y"  and  a  branch  track  may 
also  be  used  for  reversing  the  direction  of  the 
train  on  the  main  track. 

Problem  30.  A  branch  track  leaves  a 
straight  main  track  at  station  322  +  40.  The 
radius  of  the  branch  curve  is  762.7  feet.  Find 
the  P.O.  on  the  main  track  and  the  P.T.  on 
the  branch  track  of  a  "  Y"  curve,  of  573.7  feet 
radius,  that  will  connect  the  main  and  branch 
tracks.  A  is  at  station  322  +  40.  0  is  the 
center  of  the  branch  curve.  C  is  the  center, 
L  is  the  P.O.,  and  M  is  the  P.T.  of  the  "Y" 
curve. 

Let  a  =  central  angle  of  the  branch  track 
from  its  P.O.  to  the  P.T.  of  the  "Y,"  B  = 
central  angle  of  the  "  Y"  curve,  Rb  =  the  radius 
of  the  branch  curve  and  Ry  =  radius  of  the 
"Y"  curve. 

=  Rb  -  Ry  =  762.7  -  573.7 
"  Rb  +  Ry  "  762.7  +  573.7 

189 


1336.4 


=  0.142. 


TURNOUTS 


71 


By  the  use  of  the  sine  scale  of  the  slide  rule, 

sine  of  (90°  -•*)'-  0.142;  90°  -  a  =  8°  10' 

a  =  81°  50'. 
B  =  180°  -  a  =  180°  -  81°  50'  =  98°  10'. 

I  =  distance  from  P.O.  of  the  branch  to  the  P.O. 
of  the  "Y"  measured  along  the  main  track. 

A  7  B 


Fig.  30 


I  =  1336.4  - 


1892 


P.C.«  is  at 


2  X  1336.4 

322  +  40 

13  +  22.9 
335  +  62.9 


-.  =  1322.9  feet. 


72 


THE    SLIDE   RULE 

3438  X  100 

762.7 
81°  50'  X  100 


7°30/ 


=  450' =  7°  30', 
=  1090  feet, 


or  P.T.y  is  at  station  10  +  90  on  the  branch 
track. 

Problem  31.     Using  the  same  curves  for  the 
branch  and  "  Y"  tracks  as  in  No.  30,  but  having 


Fig.  31 

a  60°  central  angle  for  the  branch  curve  and 
the  P.T.  of  "Y"  coming  beyond  the  P.T.  of 
the  branch  and  on  the  tangent  to  the  branch 
curve  at  its  P.T.,  find  the  P.C.  on  the  main 
track  and  the  P.T.  on  the  branch  track  for  the 
"Y"  curve. 


TURNOUTS  73 

Using  letters  which  represent  the  same 
things  as  in  Problem  30, 

fin0  1  90° 

I  =  762.7  tan  ^-  +  573.7  tan  ~  , 

1  Z 

I  =  762.7  tan  30°  +  573.7  tan  60°, 

^70  7 
Z  =  762.7  tan  30'  +  ^, 

I  =  441  +  994  =  1435  feet. 

Settings.  Reverse  the  slide,  set  right  index 
of  T  scale  of  the  slide  opposite  762.7  on  the 
D  scale  and  opposite  30°  on  the  T  scale  read 
441  on  the  D  scale. 

Set  30°  on  the  T  scale  opposite  573.7  on 
the  D  scale,  and  opposite  the  right  index  of 
the  T  scale  read  994  on  the  D  scale. 

Let  d  =  the  distance  from  the  P.T.  of  the 
branch  curve  to  the  P.T.  of  the  "Y"  curve 
measured  along  the  tangent  to  the  branch 
curve  thru  its  P.T. 

120°       _         60° 
d  =  Ry  tan  —=-  —  Rb  tan  -_- 


-  762.7  tan  30°  =  994  -  441 


tan  30' 
553  feet. 


74  THE    SLIDE    RULE 

322  +  40 

I  =  14  +  35 

P.C.y  at  336  +  75  on  the  main  track. 

3438  X  100       ._,      ?0  ory 
762.7 

Lb  =  ^5^,  X  100  =  800  feet,  8  +  00 

d  =  5  +  53 

P.T.^at  13  +  53 
on  the  branch  track. 


CHAPTER  VI 

THE  EASEMENT  CURVE 

27.  The  purpose  of  the  easement  curve, 
connecting  the  tangent  and  the  circular  arc, 
is  to  produce  a  gradually  increasing  centrifugal 
force  that  may  be  balanced  by  a  gradually 
increasing  centripetal  force  produced  by  the 
gradual  elevation  of  the  outer  rail  of  the  ease- 
ment curve  part  of  the  track.  The  ordinary 
form  of  the  easement  curve  is  the  spiral  or  a 
similar  curve. 

From  the  above  it  is  seen  that  the  length  of 
the  spiral  is  the  distance  in  which  the  total 
amount  of  the  elevation  of  the  outer  rail  is 
gained.  In  the  best  practice  the  rate,  by 
which  the  elevation  of  the  outer  rail  is  gained, 
is  a  function  of  the  speed  of  the  train.  It  has 
been  found  that  a  gain  of  1£  inches  per  second 
is  not  felt  by  a  passenger  in  the  train.  Even 
a  gain  of  two  inches  per  second  does  not 
produce  a  disagreeable  effect. 

Let  e  =  the  elevation  of  the  outer  rail,  in 
inches,  necessary  for   the  cir- 
cular curve. 
75 


76  THE    SLIDE    RULE 

Let  v  =  the  velocity  of  the  train  in  feet 

per  second. 
S  =  the  velocity  of  train  in  miles  per 

hour. 

lc  =  the  length  of  the  spiral. 
DC  and  Rc  =  the  degree  and  the  radius  of  the 

circular  curve. 
r  =  the  rate  at  which  e  is  gained  in 

inches  per  second. 
C  =  the  centrifugal  force  of  a  car  on 

the  circular  curve. 
W  =  the  weight  of  the  car. 
G  =  the  gage  of  the  track. 
g  =  the  acceleration  due  to  gravity. 

Then  from  mechanics, 


gRc' 

From  Fig.  32, 

C       ED      ED 


Wv* 

gRc      e  Gv2 

-W=G  or  e  =  Wc'    '   * 

v  =  1.47  S,         G  =  4.71, 
g  =  32.2,   and  Rc  = 


THE    EASEMENT    CURVE  77 


Substituting  these  values  in  Equation  (21), 
reducing  and  expressing  the  result  in  inches, 
e  =  0.00067  S2DC.  At  a  rate  of  gain,  for  e,  of 
r  inches  per  second, 

I  =  eXv  =  °'WW7S*DCv 
c      r  r 

Substituting  for  v  its  value  in  terms  of  S, 

(22) 


,       0.00067  S*DC  w  1  _  0        S3DC    .  . 

lc  =  -    —f-    -  X  1.47  AS  =  1QQO  r  (approx.). 


For  a  rate  of  1£  inches  per  second, 


For  a  rate  of  2  inches  per  second, 

7    _^£ 

0  ~  2000' 


78  THE   SLIDE   RULE 

28.  The  equation  for  a  true  spiral  is  derived 
as  follows: 

If  I  is  the  distance  from  the  beginning  of 
the  spiral,  P.S.,  to  any  point,  P,  on  the  spiral 
where  its  radius  is  R  and  its  degree  of  curva- 
ture is  D, 

eP  =  0.00067  S*D 
and 

0.00067  S2X  5730  v      5.735 
~ 


r 
In  a  similar  way  it  is  readily  shown  that 


Hence 

El  =•-  Rclc  or  y  =  ^-c.    .     (23) 

'  'c 

This  is  the  equation  for  a  true  spiral  and  should 
be  used  for  very  long  curves  which  partake 
more  of  the  nature  of  a  curve  compounded 
many  times  than  of  an  easement  curve  which 
connects  a  tangent  and  a  circular  arc.  This 
equation  shows  that  the  radius  of  the  spiral 
varies  inversely  as  the  distance  from  the 
beginning  of  the  spiral,  or  its  degree  of  curva- 


THE   EASEMENT   CURVE 


79 


ture  varies  uniformly  with  the  distance  from 
the  beginning  of  the  spiral. 


B      H 
Fig.  33 

In  Fig.  33,  ADE  represents  part  of  an  ease- 
ment curve  or  spiral,  beginning  at  A,  the  point 
of  spiral.  Let  I  =  length  of  spiral  from  A  to  E, 
SE  =  the  spiral  angle  for  point  E  and  iE  =  its 
deflection  angle  from  tangent  AK. 

29.  To  find  the  value  of  SE. 

From  the  circular  curve  EF  of  radius  R  and 
degree  D, 


*B  = 


2X 


I 


D°(EF=  2>aPProx-)- 
5730 


2  X  100 
° 


R 


(24) 


80  THE    SLIDE    RULE 

Equation  (24)  gives  the  value  of  SE  in  degrees. 
In  length  of  arc  of  unit  radius 


(25) 


From  Equation  (23)  R=; 
substituting  this  value  of  R  in  Equation  (25) 


30.  To  find  the  relation  between  deflection 
angles  from  the  tangent  at  the  point  of  spiral 
to  points  on  spiral. 

Assuming  that  the  part  of  the  spiral  from 
A  to  E  is  a  circular  arc  of  D°  curvature  (prac- 
tically true  if  AE  is  a  very  small  part  of  the 
spiral),  the  deflection 

*-        or  iE=D°-  -  (27) 


D° 

The  curvature  of  the  spiral  at  point  D  =  -*-> 

z 

where  D°  is  the  curvature  of  the  spiral  at  E 

,   ,n      AE      I 
and  AD  =  —  «§. 

Considering  the  spiral  as  a  circular  curve  of 

D° 

-Tj-  from  A  to  Z), 


THE   EASEMENT   CURVE  81 


L     ®1 

22 


(28; 


-- 

IE      200  4  ,00v 

hence          •£  =  -?  -  =  T  .....     (29) 
to        *    no      1 
800^ 

Hence  deflections  from  the  tangent  at  the 
P.S.  to  points  on  the  spiral  vary  as  the  squares 
of  the  distances  from  the  P.S.  to  the  points 
on  the  spiral. 

31.  To  find  the  relation  between  the  offsets 
from  the  tangent  thru  the  P.S.  to  points  on 
the  spiral. 

pi  =  DB  =  ^  sin  iD, 

pz  =  EK  =  I  sin  iE, 
£2  =  I  sin  iE 

Pi      I   .     . 
gSiniD 

Since  the  sines  and  arcs  for  small  angles  are 
proportional,—  =  -r-^and  substituting  the  value 

of  -r*  from  Equation  (29), 
ID 

-  —  <30) 


82  THE   SLIDE    RULE 

Hence  offsets  from  the  tangent  thru  the  P.S. 
to  points  on  the  spiral  vary  as  the  cubes  of 
the  distances  to  the  said  points  from  the  point 
of  spiral. 

32.  To  find  the  value  of  the  deflections  from 
the  tangent  thru  the  P.S.  to  points  on  the  spiral. 

Since  the  spiral  changes  in  curvature  uni- 
formly with  the  distance,  the  amount  that  it 
will  deflect  from  the  tangent  for  the  distance 

AD  =  -  is  the  same  that  it  deflects  from  the 

& 

circular  arc  for  the  distance  EF  =  •=  or  FD 

Zi 

=  DB. 

Hence  F£  =  2DB  =  2P1  =  £-2  • 
From  the  circular  curve  FE, 

2 


or 

P    _3p2 

8R^   4   ' 

then 

4P 


'    ' 


=  i  sin  IE* 


THE   EASEMENT   CURVE  83 

For  small  angles  sines  and  arcs  of  unit  radii 
are  equal,  then 


or  **  =  O  ......     (32) 

j-,  _  Rclc 

T 

hence  **  = 


P 


,  .        SE 

hence  ^E  =  -$-• 

o 

Since  ^7  may  be  taken  as  any  point  on  the 
spiral, 


33.   To  show  P  =  24TR> 

p  =  FB  =  ^,    but  p2  =  /^from  Eq.  (31), 

P 

hence  p  =          .....     (35) 


34.   To  find  a  value  for  the  deflection  angle, 
i,  to  any  point  on  the  spiral. 


84  THE   SLIDE   RULE 

Let  I  =  distance  from  P.S.  to  the  point  where 

the  deflection  is  i. 
N  =  number  of  chords  from  P.S.  to  the 

same  point. 
C  =  rate  of  change  in  curvature  of  the 

spiral  per  chord  length. 
i  =  INC  X  O.I7*  giving  i  in  minutes. 

From  Equation  (32) 

I      D      5730  f 
1  =  OR '          ~~D~  (aPProxlmately)- 

*  From  Kellogg' s  Transition  Curve  by  N.  B.  Kellogg, 
C.  E. 

NOTE.  If  a  spiral  of  six  chords  in  length  is  always 
used  the  deflection  for  the  end  of  the  first  chord  is 

•|  X  1  X  -£• X  0.1  in  minutes  =  ^  -|  Dc  or  in  seconds 
oo  oU  o 

is  IN  DCJ  where  IN  is  the  length  of  a  chord,  which,  ex- 
pressed as  a  rule,  is: 

RULE.  For  a  six-chord  spiral  the  deflection  for  the 
end  of  the  first  chord  in  seconds  is  the  length  of  the 
chord  in  feet  multiplied  by  the  degree  of  the  circular 
curve. 

Deflections  for  the  ends  of  the  other  chords  may  be 
found  by  the  rule  that  the  deflections  vary  as  the  square 
of  the  distances  from  the  point  of  spiral;  i.e.,  the 
deflection  for  the  end  of  the  second  chord  is  four  times 
that  for  the  end  of  the  first  chord  and  for  the  end  of  the 
third,  nine  times  that  for  the  end  of  the  first,  etc. 


THE    EASEMENT   CURVE  85 

Hence 

ID 


6  X  5730' 
hence 

INC 


i  = 


6  X  5730 

in  length  of  arc  of  unit  radius,  and  multiplying 
by  57.3  gives  i  in  degrees, 

INC 


i  = 


6  X  100 


Multiplying  this  value  of  i  by  60  gives  i  in 
minutes. 

i  =  INC  X  0.1.     ...     (36) 

In  Equation  (36)  I  is  expressed  in  feet,  N  in 
chord  lengths  and  C  in  degrees.  N  may  have 
a  fractional  value. 

35.  To  find  the  distance  AB  in  Fig.  34, 

AB  =  q  =  ^  (approximately). 
& 

The  following  gives  a  value  for  q  more  nearly 
correct. 

In  any  right-angled  triangle  the  approximate 
difference  between  the  hypotenuse  and  base 
is  equal  to  the  square  of  the  altitude  divided 


86 


THE    SLIDE   RULE 


p.r. 


P.  S. 


by  twice  the  known  side,  either  base  or  hy- 
potenuse, hence 


Then 


1        P2 

=  2~n (37) 


P=^V      (Equation  (35)). 

l_          V 
2      4) 


....     (38) 


THE   EASEMENT   CURVE  87 

36.  To  find  the  distance  of  the  P.S.  from  V, 
the  intersection  of  the  tangents. 

Ordinarily  the  same  spiral  will  be  used  at 
each  end  of  the  circular  curve.  The  distance 
from  P.S.  to  V  will  be  found  under  this  condi- 
tion. 

AV  =  AB  +  BH  +  HV.    Let  T8  =  AV. 


T8  =  g  +  flctan    +  Ptan.    .     .     (39) 

37.  To  find  a  tangent  at  any  point  on  the 
spiral.     In  Fig.  33,  the  angle  AEH  =  EHK  - 
EAR. 

AEH=sE-iE=3iE-iE  =  2iE.     .     (40) 

Set  up  the  transit  at  E  and  with  instrument 
set  at  0°00',  sight  to  A.  Transit  the  tele- 
scope and  turn  off  an  angle  of  2  iE,  in  the 
same  direction  as  the  curve  is  running,  and 
the  line  of  sight  will  be  in  a  tangent  to  the 
spiral  at  E. 

38.  To  find  the  deflection  from  a  tangent  at 
any  point  on  the  spiral  to  any  other  point  on 
the  spiral. 

1°.  In  the  direction  from  the  P.S.  toward 
the  circular  curve. 

Let  KFL  be  an  arc  of  a  circle  of  the  same 
radius  as  that  of  the  spiral  at  F. 


88  THE   SLIDE   RULE 

Since  the  spiral  changes  in  curvature  uni- 
formly with  the  distance,  the  angular  amount 
that  the  spiral  deflects  from  the  curve  FL  is 
the  same  that  it  deflects  from  the  tangent  AB 
in  a  distance  «qual  to  FL,  or  the  angle  EFL 
equals  the  deflection  from  the  tangent  thru  the 
point  of  spiral  for  a  distance  of  FL.  The  angle 
LFM  is  the  deflection  angle  for  a  chord  of  FL 
for  the  circular  curve  of  radius  R.  The  angle 
EFM  =  EFL  +  LFM. 


Then  the  deflection  angle  from  the  tangent 
at  any  point  on  the  spiral  to  any  other  point 
on  the  spiral  is  equal  to  the  sum  of  the  deflec- 
tions for  the  spiral  and  for  the  circular  arc  for 
the  length  of  chord  between  the  two  points  on 
the  spiral,  the  circular  arc  having  the  same 
radius  as  the  spiral  at  the  point  thru  which 
the  tangent  runs,  and  the  spiral  being  run  in 
toward  the  circular  curve  which  it  connects 
with  main  tangent. 


THE   EASEMENT    CURVE  89 

2°.  In  the  direction  from  the  circular  curve 
toward  the  tangent. 

By  a  proof  similar  to  that  just  given,  it  may 
be  shown  that  the  deflection  from  the  tangent 
at  any  point  on  the  spiral  to  any  other  point 
on  the  spiral  is  equal  to  the  difference  of  the 
deflections  for  the  circular  arc  and  the  spiral 
for  the  length  of  chord  between  the  points  on 
the  spiral,  the  circular  arc  having  the  same 
radius  as  the  spiral  at  the  point  thru  which 
the  tangent  runs,  when  the  spiral  is  run  in  from 
the  circular  curve  toward  the  main  tangent. 

39.  There  are  two  common  methods  of  lay- 
ing out  easement  curves,  viz.:  first,  by  deflec- 
tion angles,  and  second,  by  offsets  from  the 
tangent  thru  the  P.S. 

The  following  steps  give  the  deflection  angle 
method  by  the  use  of  tables  derived  from  the 
formulas  and  by  the  use  of  the  slide  rule. 

1°.  From  the  assumed  speed  of  the  train, 
the  degree  of  the  curve  and  the  rate  of  gaining 
the  elevation  of  the  outer  rail,  find  the  length 
of  the  easement  curve  by  Equation  (22)  or  from 
a  table  or  diagram  made  from  this  equation. 
The  slide  rule  readily  solves  the  equation. 

2°.  From  tables,  or  by  slide  rule,  find  values 
of  p  and  q.  By  formula  or  table  find  the  tan- 


90  THE    SLIDE   RULE 

gent  distance,  Tc,  for  the  circular  curve  for  a 
central  angle  equal  to  the  deflection  angle 
between  the  tangents.  Find  value  of  T3  in 
Equation  (39). 


From  station  of  V,  intersection  of  tangents, 
subtract  T8  expressed  in  stations  and  result  is 
the  station  P.S. 

3°.  By  the  deflections  and  their  correspond- 
ing distances  from  the  P.S.  run  in  the  spiral  as 
far  as  the  P.S.C.,  where  it  joins  the  circular 
arc  (Fig.  34). 

From  the  deflection  angle  between  the  tan- 
gents subtract  twice  the  spiral  angle,  Sc,  for 
the  spiral  used,  and  the  result  is  the  central 
angle  of  the  circular  arc,  from  which  its  length 
may  be  determined. 

Set  the  transit  up  at  P.S.C.  and  run  in  the 
circular  arc  by  deflections  from  the  tangent 
thru  the  P.S.C.  ,  to  the  P.C.S.,  where  it  joins 
the  ending  spiral.  Set  up  the  transit  at  P.C.S. 
and  run  in  the  ending  spiral  to  the  P.T.  by 
deflections  from  the  tangent  thru  the  P.C.S. 
Set  up  the  transit  at  the  P.T.  and  check  work 
by  sighting  on  the  P.C.S.  and  turning  off 
one-half  of  the  deflection  angle  for  the  P.T., 


THE   EASEMENT   CURVE  91 

and  if  line  found  runs  thru  V  the  work  is 
correct. 

40.  The  method  of  laying  out  a  circular 
curve  with  spirals  at  each  end,  the  spirals  to 
be  located  by  offsets  from  tangents,  is  as 
follows: 

See  Fig.  34. 

1°.  The  circular  curve  is  to  be  shifted  from 
the  tangents  toward  the  center,  until  the  shifted 
P.O.  is  a  distance  of  "p"  from  the  tangent. 
As  the  curve  is  moved  toward  the  center  every 
point  on  the  curve  will  move  along  a  line 
parallel  to  the  line  joining  the  center  of  the 
curve  with  its  vertex.  The  shifted  P.O.  will 


Fig.  36 


be  directly  opposite  a  point  on  the  tangent  at 

a  distance  of  p  tan  -  from  the  original  position 

2^ 

of  the  P.O.,   measured   backward   along  the 
tangent. 


92  THE    SLIDE    RULE 

Set  the  instrument  up  at  E,  the  shifted  P.O., 
and  run  in  the  circular  curve  by  deflections 
from  the  tangent  EF  parallel  to  AV. 

2°.  From  the  length  of  the  spiral  found  as 
described  in  paragraph  39,  find  the  value  of  q 
and  p.  From  station  of  original  P.O.  for  the 

circular  curve  subtract  q  +  p  tan  =  to  find  the 

Zi 

station  of  A,  the  point  of  spiral. 

The  offset  from  the  tangent  A  V  to  the  middle 

fQ 

point  of  the  spiral  is  £.     By  paragraph  31,  the 

m 

offsets  to  the  spiral  from  any  point  on  the  tan- 

fY\ 

gent  A  V  is  to  ^  as  the  cube  of  the  distance  of 

'g» 

said  point  from  A  is  to  the  cube  of  ~. 

40 

Let  AB  =  -:  and  the  offset  to  the  spiral  at  B 


3  I 
Let  AC  =  -T-    and   c  =  offset  at  (7, 

27 

then        C==P- 


AD  =  I  and  d  =  offset  at  D,  then  d  =  4  p. 


THE   EASEMENT   CURVE  93 

Find  the  location  of  A,  the  P.S.,  as  described, 
then  lay  off  on  the  tangent  AB,  AK,  AC  and 
AD. 

At  B  lay  off  6;   at  K,  | ;   at  C,  c;  at  D,  d. 

Zt 

The  ends  of  the  offsets  are  in  the  spiral. 

This  method  is  not  theoretically  correct,  as 
it  assumes  that  the  distance  from  the  P.S. 
to  any  perpendicular  to  the  tangent  is  the 
same  measured  along  the  tangent  as  along  the 
spiral.  However,  the  resulting  curve  is  prac- 
tically the  same  as  the  one  found  by  a  precise 
method. 

41.  To  connect  the  two  branches  of  a  com- 
pound curve  by  a  spiral,  find  the  difference 
in  degrees  of  curvature  between  the  two 
branches.  Then  from  the  adopted  speed,  rate 
of  elevation  of  the  outer  rail  and  the  difference 
in  degree  of  curvature,  find  the  length  of  the 
spiral.  Use  the  spiral  given  in  the  tables,  that 
is  practically  of  this  length,  or  figure  the  spiral 
deflections,  etc.,  by  the  slide  rule. 

The  number  of  chords  in  this  spiral  multi- 
plied by  the  change  in  curvature  per  chord 
must  be  equal  to  the  difference  in  the  degrees 
of  curvature  of  the  two  parts  of  the  compound 
curve. 


94  THE    SLIDE   RULE 

The  "p"  of  the  spiral  will  be  the  distance 
GH  in  Fig.  37. 
AG  =  BH  is  equal  to  "q"  of  the  spiral  used. 


Fig.  37 


Assume  that  the  curve  is  to  be  run  in  from 
F  to  E.  Let  H  be  the  original  P.C.C.  Find 
B  on  the  curve  FH,  at  a  distance  of  "q"  back 
from  H.  Set  the  transit  at  B  and  run  in  the 
spiral  by  deflections  from  the  tangent  at  By  as 
given  in  paragraph  38,  to  A.  Set  the  transit 
at  A  and  run  in  the  circular  curve  AE  by 
deflections  from  the  tangent  at  A. 

42.  In  the  tables  on  pages  100  and  102  are 
given  the  deflections  from  the  tangent  through 
the  point  of  spiral,  for  different  lengths  of 
spiral  and  different  changes  in  curvature  per 
chord  length  for  ten,  twenty,  thirty,  forty  and 
fifty  foot  chords,  and  also  the  values  of  p  and  q 
for  spirals  of  different  lengths  and  changes  in 


THE    EASEMENT   CURVE  95 

curvature  per  chord.  From  the  tables,  data 
for  running  in  the  center  line  spirals  for  street 
railway  tracks  may  be  found,  using  chords 
either  10  or  20  feet  in  length. 

43.  The  table  on  page  96  shows  the  spirals 
given  in  the  tables  on  pages  100  and  102  that 
may  be  used  for  different  degrees  of  the  cir- 
cular curves. 

A  similar  table  can  be  made  for  the  spirals  to 
be  used  with  any  circular  curve  up  to  20°. 

After  selecting  the  spiral  for  any  particular 
curve  the  deflections  for  the  spiral  may  be 
taken  directly  from  the  tables,  or  be  figured 
by  the  slide  rule. 

44.  Where  a  spiral  of  the  proper  length  for  a 
given  circular  curve  cannot  be  taken  directly 
from  the  table,  the  following  example  shows  a 
method  that  may  be  used :   It  is  desired  to  use 
a  spiral  165  feet  long  for  a  5°  30'  circular  curve. 
Take  from  the  tables  the  deflections  for  the 
spiral  150  feet  long  having  a  change  of  curva- 
ture of  1°  for  each  30'  chord.     This  leaves  only 
the  deflection  for  the  end  of  the  spiral  to  be 
determined.     This  may  be  found  by  formula 
(36),  i  =  INC  X  0.1.      In  this  case  I  =  165', 
N  =  5J  and  C  =  1.     i  =  1°  30.75'. 

For  ease  in  computation,  the  last  chord  of  the 


96 


THE    SLIDE    RULE 


Degree  of 
circular 
curve. 

Length  of 
spiral. 

Change  of  curvature 

Per  30' 
chord. 

Per  40' 
chord. 

Per  50' 
chord. 

3°  00' 

tt 
ft 

tt 
it 
u 
u 
n 
u 
tt 

4°  00' 

it 

tt 
it 
n 
tt 
tt 
tt 
n 
tt 
n 

5°  00' 

tt 

tt 
tt 
tt 
tt 
tt 
ti 
tt 

60' 
80' 

90' 
100' 
120' 
160' 

180' 
200' 
240' 
300' 
60' 
80' 
100' 
120' 
160' 
180' 
200' 
240' 
300' 
320' 
400' 
90' 
120' 
150' 
160' 
200' 
250' 
300' 
400' 
500' 

1°30' 

1°30' 

1°00' 

1°30' 

0°45' 
"6°  30'"' 

1°00' 
0°45' 

0°45' 
"6°  30' 

0°30' 

2°  00' 

2°  00' 

2°  00' 

1°00' 

1°00' 

0°40' 

1°00' 
"6°  40'"' 
"0°30' 

0°30' 

0°40' 

0°30' 

1°40' 
1°15' 
1°00' 

1°40' 

1°40' 

1°15' 
1°00' 

1°15' 
1°00' 

0°30' 

0°30' 

0°30' 

THE   EASEMENT   CURVE  97 

spiral  should  be  taken  some  even  fractional  part 
of  a  full  chord  length  as  J,  J,  \  or  J  of  it.  The 
length  of  the  spiral  expressed  in  chord  lengths 
multiplied  by  the  change  in  curvature  per 
chord  length  must  always  equal  the  degree  of 
the  circular  curve. 

45.  The  following  problem  shows  the  method 
of  obtaining  by  use  of  the  tables  the  necessary 
quantities  to  locate  a  spiral  by  deflections. 

Problem  32.  Given  two  tangents  intersecting 
at  station  187  +  40,  with  a  deflection  angle  of 
33°  40',  to  find  the  stations  of  the  P.S.  and  the 
P.T.,  and  the  deflections  for  a  4°  curve,  with 
equal  spirals  at  each  end  to  connect  the  given 
tangents;  the  speed  of  train  to  be  40  miles  per 
hour  and  the  rate  of  superelevation  to  be  1.6" 
per  second.  The  length  of  the  spiral,  from 

S3D 
the  formula  lc  =  innn    ,  is  160  feet. 

1UUU  T 

The  curve  selected  from  the  tables  may  be 
either  a  160-foot  spiral  consisting  of  four  40' 
chords  with  a  change  of  curvature  of  1°  per 
chord  or  a  180-foot  spiral  consisting  of  six  30' 

2° 

chords  with  a  change  of  curvature  of  -^  per 

o 

chord.  Assuming  that  the  180'  spiral  is  used, 
the  following  is  the  solution  for  the  various 
quantities: 


98  THE   SLIDE   RULE 


Tc  =  +  o.08  =  .     .     .    .  433.5 

p  tan  i  /     =  0.94  tan  16°  50' 
log      0.94  =  9.973128  -  10 
log  16°  50'  =  9.480801  -  10 

9.453929  -  10  log  of  0.28 
.  90.0 


523.8*  feet 
V  at  station  187  +  40 

5  +  23.8 
P.S.  at  station         182  +  16.2 

1  +  80 

P.S.C.  at  station      183  +  96.2 
Deflections  from  P.S.  to  P.S.C.,  for  end  of 
each  chord  30  feet  long,  may  be  taken  from  the 
table.     For  other  points  deflections  may  be 
found  from  the  formula  i  =  INC  X  0.1'. 

Deflection  for  station  183  on  the  spiral  is 
found  as  follows: 

Distance  from  P.S.  to  183  =  83.8'. 
Distance  in  chords  of  30'  =  2.79. 

i  =  83.8  X  2.79  X  f  X  0.1  =  0°  15.6'. 

Ic  the  amount  of  curvature  in  the  circular 
arc       =I-2SC. 

Ic  =  33°  40'  -  7°  12'  =  26°  28'. 

*  Results  need  be  found  only  to  nearest  j^  of  a  foot. 


THE   EASEMENT   CURVE 


99 


X  100  =  661.7  feet. 


P.S.C.  is  at  183  +  96.2 
6  +  61.7 

P.C.S.  is  at  190  +  57.9 
1  +  80.0 

P.T.     is  at      192  +  37.9 


Stations. 

Deflections. 

Description  of  curve. 

182+16  2  P  S 

D  =  4°R 

-j-46  2        

0-02' 

7=33°  40' 

+76  2 

0-08 

Tc  =433.5' 

183                             

0-15.6 

T=523.8' 

+06  2         

0-18 

Spiral  is 

+36  2  

0-32 

180'  long 

+66  2 

0-50 

p  =  0.94' 

+96  2  P.S.C  

1-12 

g=90.0' 

184          

0-04.6 

£c=3°36' 

185 

2-04.6 

7C=26°28' 

186                         

4-04.6 

Lc=661.7' 

187                 

6-04.6 

188  

8-04.6 

189                              .     . 

10-04.6 

190                          

12-04.6 

190+57.  9  P.C.S  

13-14 

190x87  9 

0-34 

191                  

0-46.6 

+17  9 

1-04 

+47  9 

1-30 

+77.9.               .     ... 

1-52 

192  

2-05.6 

+07.9  

2-10 

37.  9  P.T. 

2.24 

100 


THE   SLIDE  RULE 


CHORD  LENGTH 

Change  of  curvature 


1°00' 

2°  00' 

3°  00' 

5°  00' 

8°  00' 

10°  00' 

1°00' 

I 

Deflections. 

10 

0-01 

0-02 

0-03 

0-05 

0-08 

0-10 

P 

Q 

20 

0-04 

0-08 

0-12 

0-20 

0-32 

0-40 

0.01 

10 

30 

0-09 

0-18 

0-27 

0-45 

1-12 

1  30 

0.02 

15 

40 

0-16 

0-32 

0-18 

1-20 

2-08 

2-40 

0.05 

20 

50 

0-25 

0-50 

1-15 

2-05 

3-20 

4-10 

0.09 

25 

60 

0-36 

1-12 

1-48 

3-00 

4-48 

6-00 

0.16 

30 

CHORD  LENGTH 

Change  of  curvature 


1°00' 

2°  00' 

3°  00' 

5°  00' 

8°  00' 

10°  00' 

1°00' 

I 

Deflections. 

20 

0-02 

0-04 

0-06 

0-10 

0-16 

0-20 

P 

Q 

40 

0-08 

0-16 

0-24 

0-40 

1-04 

1-20 

0.02 

20 

60 

0-18 

0-36 

0-54 

1-30 

2-24 

3-00 

0.08 

30 

80 

0-32 

1-04 

1-36 

2-40 

4-16 

5-20 

0.19 

40 

100 

0-50 

1-40 

2-30 

4-10 

6^0 

8-20 

0.36 

50 

120 

1-12 

2-24 

3-36 

6-00 

9-36 

12-00 

0.63 

60 

140 

1-38 

3-16 

4-54 

8-10 

13-04 

16-20 

1.00 

70 

CHORD  LENGTH 

Change  of  curvature 


30' 

40' 

45' 

1°00' 

1°15' 

1°30' 

1°40' 

2°  00' 

30' 

I 

Deflection  angles. 

30 

0°01.5 

0°-02 

0°-02.25 

0°-03' 

0°03.75 

0°04.5 

0°05' 

0°06' 

P 

q 

60 

0-06 

0-08 

0-09 

0-12 

0-15 

0-18 

0-20 

0-24 

0.03 

30 

90 

0-13.5 

0-18 

0-20.25 

0-27 

33.75 

0-40.5 

0-45 

0-54 

0.09 

45 

120 

0-24 

0-32 

0-36 

0-48 

1-00 

1-12 

1-20 

1-36 

0.21 

60 

150 

0-37.5 

0-50 

0-56.25 

1-15 

1-33.75 

1-52.5 

2-05 

2-30 

0.41 

75 

180 

0-54 

1-12 

1-21 

1-48 

2-15 

2-42 

3-00 

3-36 

0.71 

90 

210 

1-13.5 

1-38 

1-50.25 

2-27 

3-03.75 

3-40.5 

4-05 

4-54 

1.12 

105 

240 

1-36 

2-08 

2-24 

3-12 

4-00 

4-48 

5-20 

6-24 

1.68 

120 

270 

2-01.5 

2-42 

3-02.25 

4-03 

5-03.75 

6-04.5 

6-45 

8-06 

2.40 

135 

300 

2-30 

3-20 

3-45 

5-00 

6-15 

7-30 

8-20 

10-00 

3.27 

150 

THE   EASEMEN*T,.CUHVE 


Oiifcj 


10  FEET, 
per  chord. 


2°  00' 

3°  00' 

5°  00' 

8°  00' 

10°  00' 

p  and  q. 

P 

« 

P 

a 

P 

q 

P 

q 

P 

q 

0.01 

10 

0.02 

10 

0.03 

10 

0.05 

10 

0.06 

10 

0.04 

15 

0.06 

15 

0.10 

15 

0.16 

15 

0.20 

15 

0.09 

20 

0.14 

20 

0.23 

20 

0.37 

20 

0.46 

20 

0.18 

25 

0.27 

25 

0.45 

25 

0.73 

25 

0.91 

25 

0.32 

30 

0.48 

30 

0.80 

30 

1.27 

30 

1.59 

30 

20  FEET, 
per  chord. 


2°  00' 

3°  00' 

5°  00' 

8°  00' 

10°  00' 

p  and  q. 

P 

Q 

P 

q 

P 

q 

P 

q 

P 

q 

0.05 

20 

0.07 

20 

0.12 

20 

0.19 

20 

0.23 

20 

0.16 

30 

0.24 

30 

0.39 

30 

0.63 

30 

0.78 

30 

0.37 

40 

0.56 

40 

0.93 

40 

1.49 

40 

1.86 

40 

0.73 

50 

1.09 

50 

1.82 

50 

2.91 

50 

3.64 

50 

1.25 

60 

1.88 

60 

3.13 

60 

5.02 

59.9 

6.27 

59.9 

2.00 

70 

3.00 

70 

5.00 

70 

8.00 

69.9 

10.00 

69.8 

30  FEET, 
per  chord. 


40' 

45' 

1°00' 

1°15' 

1°30' 

1°40' 

2°  00' 

p  and  q. 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

0.03 

30 

0.04 

30 

0.05 

30 

0.07 

30 

0.08 

30 

0.08 

30 

0.10 

30 

0.12 

45 

0.13 

45 

0.18 

45 

0.22 

45 

0.27 

45 

0.29 

45 

0.35 

45 

0.28 

60 

0.31 

60 

0.42 

60 

0.52 

60 

0.63 

60 

0.70 

60 

0.85 

60 

0.55 

75 

0.61 

75 

0.82 

75 

1.02 

75 

1.22 

75 

1.36 

75 

1.63 

75 

0.94 

90 

1.06 

90 

1.41 

90 

1.76 

90 

2.12 

90 

2.35 

90 

2.82 

90 

1.50 

105 

1.68 

105 

2.28 

105 

2.65 

105 

3.36 

104.9 

3.74 

104.9 

4.48 

104.8 

2.24 

120 

2.51 

120 

3.35 

120 

4.16 

119.9 

5.00 

119.9 

5.53 

119.9 

6.59 

119.8 

3.19 

135 

3.60 

134.9 

4.80 

134.9 

5.95 

134.9 

7.18 

134.8 

7.95 

134.8 

9.52 

134.7 

4.36 

149.9 

4.89 

149.9 

6.51 

149.9 

8.13 

149.8 

9.77 

149.7 

10.4 

149.6 

13.0 

149.4 

&LIDE    RULE 


CHORD  LENGTH 

Change  of  curvature 


30' 

40' 

45' 

1°00' 

1°15' 

1°30' 

1°40' 

2°  00' 

30' 

40' 

I 

Deflection  angles. 

40 

0-02 

0-02.7 

0-03 

0-04 

0-05 

0-06 

0-06.7 

0-08 

P 

Q 

P 

80 

0-08 

0-10.7 

0-12 

0-16 

0-20 

0-24 

0-26.7 

0-32 

0.05 

40 

0.06 

120 

0-18 

0-24 

0-27 

0-36 

0-45 

0-54 

1-00 

1-12 

0.16 

60 

0.21 

160 

0-32 

0-42.7 

0-48 

1-04 

1-20 

1-36 

1-46.7 

2-08 

0.37 

80 

0.50 

200 

0-50 

1-06.7 

1-15 

1-40 

2-05 

2-30 

2-46.7 

3-20 

0.73 

100 

0.97 

210 

1-12 

1-36 

1-48 

2-24 

3-00 

3-36 

4-00 

4-48 

1.26 

120 

1.68 

280 

1-38 

2-10.7 

2-27 

3-16 

4-05 

4-54 

5-26.7 

6-32 

2.00 

140 

2.64 

3202-08 
36012-42 

2-50.7 
3-36 

3-12 
4-03 

4-16 
5-24 

5-20 
6-45 

6-24 
8-06 

7-06.7 
9-00 

8-32 
10-48 

2.98 
4.25 

160 
180 

3.97 
5.65 

400J3-20 

4-26.7 

5-00 

6-40 

8-20 

10-00 

11-06.7 

13-20 

5.80 

199.9 

7.70 

CHORD  LENGTH 

Change  of  curvature 


1 

30' 

40' 

45' 

1°00' 

1°15' 

1°30' 

1°40' 

2°  00' 

30' 

Deflection  angles. 

50 

0°-02'.5 

0°-03.5 

0°-03.75 

0-05 

0°-06.25 

0.07.5 

0-08.5 

0-10 

P 

<l 

100 

0-10 

0-13.5 

0-15 

0-20 

0-25 

0-30 

0-33.5 

0-40 

0.07 

50 

150 

0-22.5 

0-30 

0-33.75 

0-45 

0-56.25 

1-07.5 

1-15 

1-30 

0.25 

75 

200 

0-40 

0-53.5 

1-00 

1-20 

1-40 

2-00 

2-13.5 

2-40 

0.58 

100 

250 

1-02.5 

1-23.5 

1-33.75 

2-05 

2-36.25 

3-07.5 

3-28.5 

4-10 

1.14 

125 

30!) 

1-30 

2-00 

2-15 

3-00 

3-45 

4-30 

5-00 

6-00 

1.97 

150 

r>() 

2-02.5 

2-43.5 

3-03.75 

4-05 

5-06.25 

6-07.5 

6-48.5 

8-10 

3.12 

175 

400 

2-40 

3-33.5 

4-00 

5-20 

6-40 

8-00 

8-53.5 

10-40 

4.67 

200 

450 

3-22.5 

4-30 

5-03.75 

6-45 

8-26.25 

10-07.5 

11-15 

13-30 

6.65 

225 

,500 

4-10 

5-33.5 

6.15 

8-20 

10-25 

12-30 

13-53.5 

16-40 

9.04 

250 

THE   EASEMENT   CURVE 


103 


40  FEET, 
per  chord. 


40' 

45' 

TOO' 

1°15' 

1°30' 

1°40' 

2°  00' 

p  andg. 

q 

P 

q 

P 

q 

P 

Q. 

P 

q 

P 

q 

P 

Q. 

40 

0.07 

40 

0.09 

40 

0.12 

40 

0.14 

40 

0.16 

40 

0.19 

40 

60 

0.24 

60 

0.31 

60 

0.39 

60 

0.47 

60 

0.52 

60 

0.63 

60 

80 

0.56 

80 

0.75 

80 

0.93 

80 

1.12 

80 

1.24 

80 

1.49 

80 

100 

1.08 

100 

1.45 

100 

1.81 

100 

2.18,100 

2.42 

100 

2.90 

100 

120 

1.89 

120 

2.51 

120 

3.15 

120 

3.78J119.9 

4.19 

119.9 

5.02 

119.9 

140 

2.98 

140 

3.98 

139.9 

4.97 

139.9 

5.95(139.9 

6.61 

139.8 

7.95 

139.8 

160 

4.46 

159.9 

5.95 

159.9 

7.42 

159.8 

8.90 

159.8 

9.90 

159.7 

11.90 

159.6 

179.9 

6.38 

179.9 

8.50 

179.8 

10.60 

179.7 

12.70 

179.6 

14.10 

179.4 

16.90 

179.2 

199.8 

8.70 

199.8 

11.60 

199.7 

14.50 

199.5 

17.40 

199.3 

19.30 

199.0 

23.10 

198.7 

50  FEET, 
per  chord. 


40' 

45' 

1°00' 

1°15' 

1°30' 

1°40' 

2°  00' 

p  and  q. 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

P 

q 

0.10 

50 

0.11 

s<r 

0.15 

50 

0  18 

50 

0  22 

50 

0  24 

50 

0  29 

50 

0.33 

75 

0.37 

75 

0.49 

75 

0.62 

75 

0.74 

75 

0.82 

75 

0,98 

75 

0.78 

100 

0.87 

100 

1,15 

100 

1  42 

100 

1  74 

100 

1  94 

100 

2  32 

100 

1.51 

125 

1.70 

125 

2.26 

125 

2.84 

125 

3,40 

125 

3.79 

125 

4,53 

125 

2.63 

150 

2.95 

150 

3.92 

150 

4  90 

150 

5  91 

150 

6  54 

150 

7  88 

144  9 

4.15 

175 

4.68 

175 

6.22 

175 

7.80 

175 

9,35 

174.9 

10,39 

174  9 

12.41 

174  9 

6.22 

200 

6.98 

200 

9.34 

199  9 

11  61 

199  9 

13  98 

199  9 

15  4?, 

199  8 

18  43 

199  8 

8.87 

225 

10.0 

224.9 

13.35 

224.9 

16.6 

224.8 

20.0 

224.9 

22.1 

224  7 

26  5 

224.6 

12.1 

249.9 

13.6 

249.9 

18.1 

249.8 

22.6 

249.7 

27.2 

249.6 

30.2 

249.5 

36.2 

249.3 

104:  THE   SLIDE   RULE 

Deflection  for  station  190  +  87.9  is  found  as 
follows: 

Deflection  for  circular  arc  for  30'  chord 
=  30  X  0.3  X  4  =  36  minutes. 

Deflection  for  spiral  for  30'  chord  from 
P.S.  =  0°  02'. 

Deflection  from  tangent  through  station 
190  +  57.9  (P.C.S.)  =  0°  36'  -  0°  02'  =  0°  34'. 

The  same  method,  the  theory  of  which  is  de- 
veloped in  second  part  of  paragraph  38,  is  used 
to  obtain  the  other  deflections  from  P.C.S. 
to  P.T. 

The  deflections  from  P.S.  to  P.S.C.  are  from 
the  tangent  thru  the  P.S.  The  deflections  from 
P.S.C.  to  P.C.S.  are  from  the  tangent  thru  the 
P.S.C.  The  deflections  from  P.C.S.  to  P.T. 
are  from  the  tangent  thru  the  P.C.S. 

46.  The  following  problem  gives  the  method 
for  the  solution  of  easement-curve  problems  by 
the  use  of  the  slide  rule. 

Problem  33.  Find  the  deflections  for  a  3°  40' 
curve  with  spirals  at  each  end  to  connect  the 
tangents  shown,  using  a  rate  of  gain  of  eleva- 
tion of  the  outer  rail  of  2"  per  second  and  a 
speed  of  50  miles  per  hour. 

-  23° feet 


THE    EASEMENT   CURVE  105 

Setting.  Opposite  50  on  the  D  scale  set 
2000  on  the  B  scale,  place  the  runner  over  3.67 
on  the  B  scale,  bring  the  right  index  of  the  B 
scale  to  the  runner  and  opposite  50  on  the  B 
scale  read  230  on  the  A  scale,  the  decimal  point 
for  the  result  being  found  by  rough  figuring. 


Fig.  38 

Practically  the  same  result  may  be  found 
more  quickly  by  the  use  of  the  diagram  on 
page  116. 

A  six-chord  curve  of  40-foot  chords  will  give 
a  spiral  240  feet  long.  This  length  will  be  used, 
as  it  is  in  accordance  with  ordinary  practice, 
but  it  will  be  shown  that  the  230-foot  spiral 
can  be  almost  as  easily  solved. 

Let  P.S.  be  the  beginning  of  the  spiral  at  the 

tangent  end, 
P.S.C.  be  the  ending  of  the  spiral  at  the 

circular  curve  end, 
P.C.S.  be  the  beginning  of  the  ending 

spiral  at  the  circular  curve  end, 


106  THE    SLIDE    RULE 

P.T.  be  the  end  of  the  ending  spiral, 
n  be  the  number  of  chords  in  the  spiral, 
I  be  the  chord  length  in  feet, 
i\  be  the  deflection  from  the  tangent  at 
the  P.S.  to  the  end  of  the  first  chord. 

Then 

it  =  0.1  X  I  -  =  0.1  X  40  ^  =  2.44  minutes. 
n  6 

Setting.  Opposite  4  on  the  D  scale  set  6  on 
the  C  scale  and  opposite  3.67  on  the  C  scale 
read  2.44  on  the  D  scale. 

For  the  station  of  the  P.S.  subtract  from  the 
station  of  the  vertex  the  sum  of  the  tangent 
distance  of  the  circular  curve  of  central  angle  of 
32°  40',  the  half  length  of  the  spiral  and  the 
spiral  offset  multiplied  by  the  tan  16°  20',  or 


T8  = 


220 

Q  = 

120     feet. 

2402 

2402               M  f 

*ft* 

1.54  tan  16°  20'=  0.5  feet. 

T8  =  577.5  feet. 


THE   EASEMENT   CURVE  107 

V  is  at  station  413 

5  +  77.5 

P.S.  is  at  station       407  +  22.5 
lc    =  2  +  40 

P.S.C.  is  at  station  409  +  62.5 

The  deflections  for  the  spiral  vary  as  the 
squares  of  the  distances  from  the  P.S.  Hence 
the  deflection  for  the  end  of  the  spiral  is 

i6  =  62'ii  =  62  X  2.44  =  87.75'  =  1°  27.75' 
s6  =  3  ig  =  4°  23'. 

As  there  are  two  spirals  the  total  amount 
of  central  angle  used  up  by  the  spirals  is  2  X 
4°  23'  =  8°  46',  leaving  32°  40'  -  8°  46'  =  23°  54' 
in  the  central  angle  of  the  circular  part  of  the 
curve. 

Let          lc  —  length  of  the  circular  part 

23°  54' . 
=   go  4Q,  m  stations 

=  -SrtTr  =  6-52  stations. 

ZZ(J 

P.S.C.  409  +  62.5 

Lc  6  +  52 

P.C.S.  416  +  14.5 

lc  2  +  40 

P.T.  418  +  54.5 


108 


THE    SLIDE   RULE 


Station. 

Deflection. 

Station. 

Deflection. 

407+22.5 

P.S. 

409+62.5 

P.S.C. 

+62.5 

0°02.44/ 

410 

0°41' 

408 

0°09.2' 

411 

2°  31' 

+02.5 

0°09.S' 

412 

4°  21' 

+42.5 

0°22' 

413 

6°  11' 

+82.5 

0°39' 

414 

8°  01' 

409 

0°48' 

415 

9°  51' 

+22.5 

1°01' 

416 

11°  41' 

+62.5 

1°28' 

+14.5 

11°  57' 

Station. 

Deflection. 

416+14.5 

P.C.S. 

+54.5 

0°41.6' 

+94.5 

1°  18.2' 

417 

1°24' 

+34.5 

1°52' 

+74.5 

2°  17' 

418 

2°  31.  5' 

+14.5 

2°  39' 

+54.5  P.T. 

2°  56' 

Number  of  chords  from 

Sta.  407  +  22.5  to  Sta.  408  = 


77.5 
40 


Sta.  407  +  22.5  to  Sta.  409  = 


40 


=  1.94, 


=  4.44. 


The  deflections  for  the  first  spiral  are  found 
from  the  following  setting:     Opposite  2.44  on 


THE   EASEMENT   CURVE 


109 


the  A  scale  set  the  index  of  the  slide  and 
opposite  1.94,  2,  3,  4,  4.44,  5  and  6  on  the  C 
scale  read  9.2,  9.8,  22,  39,  etc.,  on  the  A  scale. 

The  setting  for  the  deflections  of  the  circu- 
lar part  has  been  given  in  a  preceding  problem. 

The  deflections  for  the  circular  part  are  from 
the  tangent  thru  the  P.S.C. 

The  deflections  for  the  ending  spiral  are  from 
the  tangent  thru  the  P.C.S.  and  are  found  by 
subtracting  the  deflection  for  the  spiral  for  the 
given  chord  length  from  the  deflection  for  the 
circular  curve  and  for  the  same  chord  length, 
e.g.,  the  deflection  for  40  feet  for  a  3°  40' 
curve  is  44'  and  for  40  feet  for  the  spiral  is  2.44', 
as  found  above;  then  the  deflection  for  the  end 
of  the  first  40-foot  chord  of  the  ending  spiral 
is  44'  -  2.4'  =  41.6'. 

The  deflections  for  the  ending  spiral  are 
found  as  follows: 


Station. 

Circular  curve. 

Spiral. 

Ending  spiral. 

416+14.5 

P.C.S. 

+54.5 

0°44' 

0°02.4' 

0°41.6' 

+94.5 

1°28' 

0°09.8' 

1°18.2' 

417 

1°34' 

0°10' 

1°24' 

+34.5 

2°  14' 

0°22' 

1°52' 

+74.5 

2°  56' 

0°39' 

2°  17' 

418 

3°  24' 

0°52.5' 

2°  31.  5' 

+14.5 

3°  40' 

i°or 

2°  39' 

+54.5 

4°  24' 

1°28' 

2°  56' 

110  THE   SLIDE    RULE 

After  the  stations  of  the  P.S.,  P.S.C.,  P.C.S. 
and  P.T.  have  been  determined,  the  skeleton 
for  the  rest  of  the  computation  can  be  laid  out 
and  all  the  deflections  for  the  circular  part, 
including  those  for  finding  the  deflections  of 
the  ending  spiral,  can  be  found  from  one  set- 
ting of  the  slide  rule.  Likewise  all  the  deflec- 
tions for  the  spirals  can  be  found  from  one 
setting  of  the  slide  rule. 

If  spirals,  each  230  feet  long  consisting  of 
five  40-foot  chords  and  one  30-foot  chord,  are 
used  in  place  of  the  above,  the  following  solu- 
tion is  made: 

T8  =  Tc  +  q 


Ts  =  457  +  115  +  |rir  tan  16°  20' 

Z4  £lc 

=  572.4  feet. 

V  at  413 
Ts=  5  +  72.4 

P.S.  at  407  +  27.6 
Ic  =  2  +  30 

P.S.C.  at  409  +  57.6 
Lc=  6  +  61 

P.C.S.  at  416  +  18.6 
Ic  =  2  +  30 

P.T.  at  418  +  48.6 


THE    EASEMENT   CURVE 


111 


402 


X  1°  24'  =  2.52'. 


2302 

Number  of  40-foot  chords  from 

70  4 
407  +  27.6  to  408  =  -^~  =  1.81. 


407  +  27.6  to  409  = 


40 


40 


=  4.31. 
230 


407  +  27.6  to  409  +  57.6  =  =^  =  5.75. 

4U 

2c*     oo  cinf 
Oc   —   O      £\J  • 

7C  =  32°  40'  -  8°  26'  =  24°  14'. 

24°  14'      1454 
Lc  =  -g^Qr  =  220"  =  6'61  Statl0ns' 


Station. 

Deflection. 

Station. 

Deflection. 

407+27.6 

P.S. 

409+57.6 

P.S.C. 

+67.6 

0°02.52r 

410 

0°46.5' 

408 

0°08.3' 

411 

2°  36.  5' 

+07.6 

0°10.4r 

412 

4°  26.  5' 

+47.6 

0°22.6/ 

413 

6°  16.5' 

+87.6 

0°40.5/ 

414 

8°  06.  5' 

409 

0°47/ 

415 

9°  56.  5' 

+27.6 

1°03' 

416 

11°  46.  5' 

+57.6 

1°24' 

+  18.6 

12°  07' 

112 


THE    SLIDE   RULE 


Station. 

Deflection. 

Circular  curve. 

Spiral. 

416+18.6 

P.C.S. 

+58.6 

0°41.f>' 

0°44' 

0°02.5' 

+98.6 

1°17.5' 

1°28' 

0°  10.4' 

417 

l°19' 

1°29.5' 

0°  10.5' 

+38.6 

1°51.4' 

2°  14' 

0°22.6' 

+78.6 

2°  15.  5' 

2°  56' 

0°40.5' 

418 

2°  27' 

3°  19' 

0°52' 

+18.6 

2°  37' 

3°  40' 

1°03' 

+48.6 

2°  48' 

4°  12' 

1°24' 

This  solution  shows  the  flexibility  of  this 
method  which  may  be  used  for  any  length  of 
spiral  with  any  unit  chord  length.  With  a 
little  experience  with  the  slide  rule  solutions 
can  be  made  more  rapidly  than  by  the  use  of 
tables. 

47.  The  following  gives  in  a  concise  form 
the  equation  for  finding  the  deflections  for  a 
spiral  and  for  which  the  use  of  the  slide  rule 
is  particularly  adapted. 

By  Equation  (36), 


Let 
Then 


0.1 

L  =  the  chord  length. 
NL  =  I 


and 


10 


(41) 


THE   EASEMENT   CURVE 


113 


The  next  table  gives  the  deflections,  for  a 
change  of  curvature  of  one  degree  per  chord, 
at  the  end  of  chords  of  10  feet  each. 

Deflections  for  chords  of  10  feet  and  for  a 
change  of  curvature  of  one  degree  per  chord. 


1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

G°or 

0°04' 

0°09' 

0°16' 

0°25' 

0°36' 

0°49' 

1°04' 

1°21' 

1°40' 

For  any  other  chord  length  and  change  of 
curvature  per  chord,  multiply  the  tabular 
amount  by  the  given  change  of  curvature,  in 
degrees,  per  chord,  and  then  multiply  this 
product  by  the  given  chord  length  divided  by 
10,  the  result  being  the  required  deflection. 

For  fractional  chords  multiply  the  deflection, 
found  for  the  first  chord,  by  the  squares  of 
the  fractions  which  represent  the  distances 
of  the  points  from  the  beginning  of  the  spiral, 
expressed  in  chords. 

48.  The  following  diagram  is  for  finding  e 
in  inches  from  the  speed  of  the  train  and  the 
degree  of  the  curve. 

In  using  it  find  the  intersection  of  the  curve 
and  the  line  of  the  adopted  speed,  then  go 
parallel  to  the  x  axis  to  the  intersection  with 
the  line  of  the  degree  of  the  curve,  then  go 


114 


THE    SLIDE    RULE 


THE    EASEMENT    CURVE  115 

parallel  to  the  y  axis  and  read  the  result  on 
the  bottom  line  of  the  diagram,  e.g.,  to  find  the 
value  of  e  for  a  speed  of  40  miles  per  hour 
and  for  a  4°  curve,  the  result  would  be  4.6 
inches. 

49.  The  next  diagram  is  for  finding  lc  in 
feet  from  the  adopted  speed  in  miles  per  hour, 
the  degree  of  the  curve  and  the  adopted  rate 
of  gaining  the  elevation  of  the  outer  rail. 

In  using  it,  find  the  intersection  of  the  line 
of  the  adopted  speed  and  the  curve,  then  go 
parallel  to  the  x  axis  to  the  intersection  with 
the  line  of  the  degree  of  the  curve,  then  go 
parallel  to  the  y  axis  to  the  intersection  with 
the  line  of  the  adopted  rate  of  gain  of  the 
elevation  of  the  outer  rail,  then  go  parallel  to 
the  x  axis  to  the  left  line  of  the  diagram  and 
read  the  length  of  the  spiral,  e.g.,  to  find  the 
length  of  the  spiral  for  a  4°  curve,  speed  40 
miles  per  hour  and  a  rate  of  gaining  the  eleva- 
tion of  the  outer  rail  of  1.6  inches  per  second, 
find  the  intersection  of  the  40  miles  per  hour 
line  with  the  curve,  then  go  parallel  to  the 
x  axis  to  the  4°  line,  then  go  parallel  to  the  y 
axis  to  the  1.6  line,  then  go  parallel  to  the  x 
axis  to  the  left  line  of  the  diagram  and  read 
160  feet. 


116 


THE    SLIDE    RULE 


-o 
Q. 


y  axis 


i 


j  ?» 

Qo 

co     O 

c^o 


bfl 

c^ 

Q 


O  O  O 

IO  O  to 

cr>         ^-          ^ 


CHAPTER  VII 

EARTHWORK 

50.  For  finding  the  volumes  of  earthwork  in 
railroad  grading,  it  is  necessary  to  take  cross- 
sections  in  the  field  and  from  the  notes  taken 
the  volumes  may  be  figured. 

The  method  of  taking  cross-sections  where 
an  ordinary  level  is  used  is  as  follows: 

1st.  From  the  grade  line  established  on  the 
profile  made  from  the  levels  run  over  the  center 
line  of  the  survey,  find  the  elevation  of  grade 
at  the  station  where  the  cross-section  is  to  be 
taken. 

2nd.  From  the  readings  taken  on  some 
bench  mark  and  on  turning  points,  find  the 
H.I.  of  the  level. 

3rd.  Subtract  the  elevation  of  grade  from 
the  H.I.  of  the  level  and  the  result  is  rg,  the 
rod  to  grade. 

4th.  Read  the  rod  held  at  the  center  line 
stake  and  subtract  this  reading,  r8,  from  rg 
and  the  result  is  the  center  cut  or  fill,  cut  if  + 
and  fill  if  -. 

5th.  Assuming  the  section  to  be  a  level 
117 


118  THE    SLIDE    RULE 

section;  figure,  from  the  given  width  of  roau- 
bed  and  ratio  of  the  side  slope,  the  distance 
from  the  center  line  to  the  edge  of  the  side 
slope  (equal  to  one-half  of  the  width  of  the  road- 
bed plus  the  side  slope  times  the  center  cut 
or  fill).  Going  out  this  distance  estimate  the 
difference  of  elevation  between  this  point  and 
the  center  line  point.  Apply  this  difference 
to  the  cut  or  fill  at  the  center  and  from  the 
resulting  amount  of  cut  or  fill  figure  anew  the 
distance  to  the  edge  of  the  side  slope.  Go  out 
this  distance  from  the  center  stake  and  on  a 
line  at  right  angles  to  the  center  line.  At  the 
point  thus  found  take  a  rod  reading.  Sub- 
tract this  reading  from  the  rg  and  the  result 
is  the  cut  or  fill.  From  this  cut  or  fill  figure 
anew  the  distance  out  from  the  center  and  if 
this  exceeds  the  actual  distance  out,  go  out 
further  and  if  less  than  the  actual  distance  go 
nearer  the  center  line,  until  a  point  is  found 
where  the  reading  of  the  rod  gives  a  cut  or 
fill  that  corresponds  with  the  distance  from 
the  center  line.  Where  the  section  is  not  level 
across  the  top,  the  three-level  form  of  section 
is  usually  found. 

51.  The  following  is  a  form  for  the  notes 
of  cross-sections: 


EARTHWORK 


119 


Station. 

Eleva- 
tion. 

Grade. 

L 

c 

R 

85 

105.3 

105.2 

0.0/10.0 

+0.1 

+0.4/10.6 

84 

107.4 

104.6 

+2.0/13.0 

+2.8 

+3.2/14.8 

83 

110.2 

104.0 

+4.2/16.3 

+6.2 

+7.0/20.5 

82 

109.4 

103.4 

+4.2/16.3 

+6.0 

+7.2/20.8 

81 

106.6 

102.8 

+3.0/14.5 

+3.8 

+4.0/16.0 

80 

107.8 

102.2 

+4.6/16.9 

+5.6 

+5.8/18.7 

52.  Where  the  grade  contour  does  not  cross 
the  center  line  at  right  angles  at  a  point  where 
there  is  a  change  from  cut  to  fill  or  vice  versa, 
Fig.  39  shows  the  points  that  must  be  taken 
in  cross-section  work. 


312     s     Center  D\  SIS  &?2~~nr~L*ne        ~         814 

p— O  "Tfl 7 oft*- -  — ^i^       "   I.  "      g  I 


Fig.  39 

D  is  at  station  312  +  60,  F  is  opposite 
station  312  +  78,  K  is  at  station  313  +  06, 
H  is  at  station  313  +  38,  and  E  is  opposite 
station  313  +  62.  The  lower  diagrams  show 
the  forms  of  the  cross-sections  at  D  and  H 
respectively. 


120 


THE   SLIDE    RULE 


From  station  312  to  D  is  in  cut  and  may 
be  figured  by  the  average  end  area  method. 
From  D  to  E  is  in  cut  and  may  be  figured  as  a 
pyramid  of  the  base  ABCD  and  of  the  height 
equal  to  the  distance  from  D  to  E  measured 
along  the  center  line.  From  F  to  H  is  in  fill 
and  may  be  figured  as  a  pyramid  of  the  base 
GHJI  and  of  the  height  equal  to  the  distance 
from  F  to  H  measured  along  the  center  line. 
From  H  to  station  314  is  in  fill  and  may  be 
figured  by  the  average  area  method. 

The  following  gives  the  notes  for  this  part 
of  the  cross-section  work: 


Station. 

Eleva- 
tion. 

Grade. 

L 

c 

R 

314 
313+62 

164.0 

166.1 

-1.8/9.7 
0.0/10.0 

-2.1 

-2.4/10.6 

313+38 
313+06 

165.7 
167.0 

166.7 
167.0 

0.0/7.0 

-1.0 
0.0 

-1.2/8.8 

313 

167.4 

167  1 

+0  3 

312+78 

0  0/7.0 

312+60 
312 

170.2 
172.4 

167.5 
168.1 

+3.4/15.1 

+5.2/17.8 

+2.7 
+4.3 

0.0/10.0 

+3.2/14.8 

53.  A  level  section  is  one  at  which  the  sur- 
face of  the  ground  at  right  angles  to  the  center 
line  is  horizontal.  The  area  of  a  level  section 
is  found  as  follows: 


EARTHWORK 


121 


Let  6  =  the  width  of  the  roadbed  in  feet, 
s  =  the  ratio  of  the  side  slope,  horizontal 

to  vertical. 
c  =  the  center  cut  or  fill  in  feet. 


Fig.  40 

From  Fig.  40,  it  is  seen  that  the  area  of  the 

section  is  equal  to  be  +  sc  X  c, 

or  A  =  (b  +  sc)c.      .     .     .     (42) 

54.  A  three-level  section  is  shown  in  Fig.  41. 
The  amounts  of  the  cut  or  fill  at  the  center 
and  the  edges  of  the  side  slopes  must  be  found. 
The  area  may  be  expressed  in  two  ways: 

1st.  By  the  use  of  the  grade  triangle.  In 
Fig.  41,  ABFDE  is  the  three-level  section  and 
BHF  is  its  grade  triangle. 

EG       b 


The  figure  AHDE  may  be  divided  into  two 
triangles  with  the  common  base  EH  =  c+^- 
and  of  altitudes  di  and  dr  respectively. 


122 


THE    SLIDE   RULE 


Area^FD£  =  (c  +  A)^_^X» 


Let 
Then 


\ 
\ 

N\j 

Fig.  41 

G              / 
H 

D 

=  dr  +  di. 

B 


G 
Fig.  42 


2nd.  Without  the  use  of  the  grade  triangle. 
In  Fig.  42,  the  area  ABFDE  is  divided  into 
four  triangles,  AEG  and  GFD  having  equal 


EARTHWORK  123 

bases,  each  equal  to  ~,  and  AGE  and  GED  hav- 

Zi 
ing  the  same  altitude  c. 

Area  ABFDE  =  %(dr  +  di)  +  \  ^y^  -     (44) 

55.  The  two  methods  most  commonly  used 
for  figuring  volumes  of  earthwork  are: 

1st.  The  average  end  area,  which  is  ob- 
tained by  simply  dividing  the  sum  of  the  end 
areas  by  two  and  multiplying  the  result  by 
the  distance  between  the  end  sections.  The 
last  result  divided  by  27  gives  the  volume  in 
cubic  yards,  if  the  measurements  given  in  the 
notes  are  in  feet. 

2nd.  The  use  of  the  prismoidal  formula  for 
precise  results.  If  A\  and  A2  are  the  end 
areas,  Am  the  middle  area  and  h  the  length  of 
the  earthwork  section,  then,  by  the  prismoidal 
formula, 

V  =  ^(Al  +  ±Am  +  A2).   .     .    (45) 

Am  may  be  found  by  taking  a  mean  of  the 
dimensions  of  the  end  sections  and  figuring  its 
area  from  these  as  its  dimensions. 

56.  By  Equation  (46)  the  prismoidal  correc- 
tion   may    be   found.     This    correction    sub- 


124 


THE    SLIDE   RULE 


tracted  from  the  volume  given  by  the  average 
end  area  method  gives  the  precise  result. 


Cp  = 


~  Da)-  •  (46) 


Cp  is  in  cubic  yards  if  ci,  Co,  DI  and  Z)0  are 
in  feet. 

In    Fig.   43,   the    volume  A\GiFiF^G^  is 
the  same,  whether  figured  by  the  average  end 


area  method  or  by  the  prismoidal  formula, 
because  GiFi  =  GoF0.  Likewise,  the  volume 
FiDiEiEoFoDQ  needs  no  correction  if  figured 
by  the  average  end  area  method. 

If  the  entire  volume  of  earthwork  between 
sections  zero  and  1  is  figured  by  the  average  end 
area  method,  then  the  correction  to  be  applied, 


EARTHWORK  125 

to  get  precise  results,  comes  from  the  parts 
o    and   B^^^^D^.     For   the 


sum  of  these  parts  the  area  at  section  1  =  -  —  , 

JU 


at  section  zero=  -^,  and  at  the  middle  sec- 

4H 

Co  +  Ci   DO  +  DI  , 

tion  =  —  ^—     0          ,  where  CQ  and  a  are 

Zi  A  /\  £ 

the  respective  center  heights  and  D0  and  DI 
are  the  respective  total  widths  of  the  sections. 
Substituting  these  values  in  the  equations  for 
the  volumes, 


VB  =     2     '     2    .  _  I 
2      ~L~\ 


I  /CjDi  Cp  +  Ci  DQ  +  DI         CoD0\ 

+;          --  +  - 


=       (2  ClZ)!  +  CoDi  +  ciA>  +  2  c0D0). 
Subtract  Vp  from  FE  for  Cp. 

Cp  = 


126  THE   SLIDE   RULE 

This  expression  shows  that  if  the  similar 
dimensions  of  the  end  section  have  about  the 
same  values,  the  average  end  area  method 
gives  close  results. 

If  I  =  100  feet  and  Cp  is  expressed  in  cubic 
yards 


=  3       (ci  -  Co)  (Di  -  A>.) 

57.  For  finding  the  amounts  of  earthwork 
from  cross-section  notes,  the  slide  rule  does 
not  give  as  rapid  solutions  as  diagrams  or 
tables.  Where  these  are  not  available  its  use 
is  advisable.  For  this  class  of  problems  the 
following  equations  are  in  forms  readily  solved 
by  the  slide  rule. 

1st.  For  volumes,  in  cubic  yards,  of  level 
sections  for  lengths  of  50  feet, 

50  ,    .  50 
F50  =  27c6  +  27sc' 

Where  c  is  the  center  cut  or  fill,  b  is  the  width 
of  roadbed  and  s  is  the  ratio  of  the  side  slope, 
horizontal  to  vertical. 

For  b  =  20  feet  and  s  =  1J, 

Fso  =  37.04  c  +  2.78  c2; 


EARTHWORK  127 

For  b  =  14  feet  and  s  =  1|, 

Foo  =  25.95  c  +  2.78  c2; 
For  b  =  33  feet  and  s  =  1^, 

F50  =  61.11  c  +  2.78c2; 

For  b  =  27  feet  and  s  =  1^, 

75o  =  50  c  +  2.78  c2. 

2nd.   For  volumes,  in  cubic  yards,  of  three 
level  sections  for  lengths  of  50  feet, 


T7       50^     ,   50    b  /rk      ,, 
V*>=UD*  +  &2-8(D-V- 

Where  A  is  the  area  of  the  section  in  square 
feet  and  D  is  the  entire  width  of  the  section, 
from  left  side  height  to  right  side  height,  and 
the  other  letters  represent  the  same  quantities 
as  in  the  first  case. 

For  6  =  20  feet  and  s  =  1J, 
ion 


For  6  =  14  feet  ands  =  1J, 
100 


128  THE    SLIDE   RULE 

For  6  =  33  feet  and  s  =  1 1, 
100 


For  6  =  27  feet  and  s=  li, 
inn 


=        Dc  +  8*33  (£- 


The  following  problems  show  the  use  of  the 
slide  rule  for  solving  earthwork  problems. 

Problem  34.  Find  the  volume,  in  cubic 
yards,  of  earthwork  from  station  116  to  sta- 
tion 118,  assuming  that  the  sections  are  level 
sections. 

6  =  14  feet.  s  =  1J. 

F50  =  25.95  c  +  2.78  c2. 


Stations. 

Elevations. 

Grade. 

c 

316 
317 
318 

315.1 
313.6 
316.8 

318.2 
319.0 
319.8 

-     3.1 
-     5.4 
-     3.0 

316 
317 
318 

80.5       +  26.6 
(140.1       +  81.0)X2 
77.9       +  25.0 

=  107.1 
=  442.2 
=  102.9 

yds. 

652.2  cu. 

Problem  35.     Find    the    volume,    in    cubic 
yards,  of  the  earthwork  from  station  212  to 


EARTHWORK 


129 


station  214,  assuming  that  the  sections  are 
level  sections. 

b  =  33  feet.  s  =  If. 

750  =  61.11c  +  2.78c2. 


Stations. 

Elevations. 

Grade. 

c 

212 
213 
214 

153.9 
150.4 
146.9 

147.2 
146.6 
146.0 

+     6.7 
+     3.8 
+    0.9 

212 
213 
214 

409.0           +124.5        =533.5 
2(232.0           +  40.0)      =544.0 
54.9           +    2.2       =  57.1 

1134. 6  cu.  yds. 


Problem  36.  Find  the  volume,  in  cubic 
yards,  of  the  earthwork  from  station  272  to 
station  274.  i 

b  =  20  feet.  s  =  1 J. 


Sta- 
tions. 

Eleva- 
tions. 

Grade. 

L 

C 

R 

272 
273 
274 

153.9 
150.4 
146.9 

147.2 
146.6 
146.0 

+  7.8 

+6.7 
+3.8 
+0.9 

+  6.0 

21.7 

+  5.4 

19.0 

+  3.2 

18.1 

+  2.2 

14.8 
+  0.6 

13.3 

10.9 

272 
273 

274 

252.5+127.5 
2(115.5+  79.5) 
20.1+  25.9 

=  380.0 
=  390.0 
=  46.0 

816.0  cu.  yds. 

130 


THE    SLIDE   RULE 


Problem  37.  Find  the  volume,  in  cubic 
yards,  of  earthwork  from  station  376  to  sta- 
tion 378. 

b  =  27  feet.  s  =  1J. 

100 
FM  =  c  +  8.33  (D  -  6). 


Sta- 
tions. 

Eleva- 
tions. 

Grade. 

L 

C 

R 

°«7fi 

Q1  K     1 

°,1<3  9 

-  1.2 

Q    1 

-4.6 

077 

01  q   (\ 

01  Q  n 

15.3 

-2.8 

C     A 

20.4 

-7.8 

070 

01  f*      Q 

01Q     0 

17.7 
-2.2 

•*  n 

25.2 
-  3.4 

0/0 

376 

OIO  .0 

102  5 

4-  72  5 

16.8 
=  175  0 

18.6 

377 

2(214  5 

+132.5) 

=  694.0 

378 

98.3 

+  70.0 

=  168.3 

1037.  3  cu.  yds. 

Problem  38.     Find  the  prismoidal  correction 
for  the  volume  found  in  Problem  36. 


Cp  = 


~  Co)  (Dl  ~ 


From  station  272  to  station  273 
2.9  X  7.8 


3.24 


=  6.98. 


EARTHWORK  131 


From  station  273  to  station  274 
2.9  X  8.7        7.78 


CP  = 


3.24        "  14.76  cu.  yds. 


Problem  39.   Find  the  prismoidal  correction 
for  the  volume  found  in  Problem  37. 
From  station  376  to  station  377 

2.3  X  7.2 
Cp-     -£24- 

From  station  377  to  station  378 

=  2.4  X  7.5  =  5.55 

3.24  10.66  cu.  yds. 

There  are  other  problems  in  curves  and 
earthwork  in  which  the  slide  rule  may  be  used 
to  advantage.  However,  a  sufficient  number 
of  problems  has  been  given  to  guide  the  stu- 
dent in  deciding  when  to  use  it. 


132 


THE   SLIDE   RULE 


05 

1 

-o 

03 

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B  I-O      U  IrO      U  I 


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FORMULAS 


133 


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d  oq 
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134 


THE    SLIDE   RULE 


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s 


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10  s  8 

03       03 


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FORMULAS 


135 


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O 


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II 

136 


THE    SLIDE    RULE 


.    I 

I  I 


•S  i 

•+3     £2- 


I  |    i 

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S     " 


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